SEBA Class 10 Advanced Maths Chapter 4.1 Solutions: Quadratic Equation

Get free SEBA Class 10 Advanced Maths Chapter 4.1 Solutions: Quadratic Equation. This article provides 100% reliable and accurate solutions for all the questions of Exercise 4.1 from the SEBA Class 10 Advanced Maths textbook for the upcoming HSLC Examination.

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Chapter 3.2 of SEBA Class 10 Advanced Mathematics

Chapter 3.3 of SEBA Class 10 Advanced Mathematics

Class 10 Advanced Maths Chapter 4.1 Solutions

Question 1: Form a quadratic equation with the following roots:

(i) $-5, 7$

Solution:

The required equation is:

$x^2 – (-5 + 7)x + (-5) \times 7 = 0$

$\implies x^2 – 2x – 35 = 0$

$\therefore x^2 – 2x – 35 = 0$

(ii) $-\frac{1}{2}, -3$

Solution:

The required equation is:

$x^2 – \left(-\frac{1}{2} + (-3)\right)x + \left(-\frac{1}{2}\right) \times (-3) = 0$

$\implies x^2 – \left(-\frac{7}{2}\right)x + \frac{3}{2} = 0$

$\implies x^2 + \frac{7}{2}x + \frac{3}{2} = 0$

$\therefore 2x^2 + 7x + 3 = 0$

(iii) $1, -\frac{3}{2}$

Solution:

The required equation is:

$x^2 – \left(1 + \left(-\frac{3}{2}\right)\right)x + 1 \times \left(-\frac{3}{2}\right) = 0$

$\implies x^2 – \left(-\frac{1}{2}\right)x – \frac{3}{2} = 0$

$\implies x^2 + \frac{1}{2}x – \frac{3}{2} = 0$

$\therefore 2x^2 + x – 3 = 0$

(iv) $1, -\frac{4}{5}$

Solution:

The required equation is:

$x^2 – \left(1 + \left(-\frac{4}{5}\right)\right)x + 1 \times \left(-\frac{4}{5}\right) = 0$

$\implies x^2 – \left(\frac{1}{5}\right)x – \frac{4}{5} = 0$

$\implies x^2 – \frac{1}{5}x – \frac{4}{5} = 0$

$\therefore 5x^2 – x – 4 = 0$

(v) $\frac{1}{2}, -\frac{1}{3}$

Solution:

The required equation is:

$x^2 – \left(\frac{1}{2} + \left(-\frac{1}{3}\right)\right)x + \frac{1}{2} \times \left(-\frac{1}{3}\right) = 0$

$\implies x^2 – \left(\frac{1}{6}\right)x – \frac{1}{6} = 0$

$\implies x^2 – \frac{1}{6}x – \frac{1}{6} = 0$

$\therefore 6x^2 – x – 1 = 0$

(vi) $5i, -5i$

Solution:

The required equation is:

$x^2 – (5i + (-5i))x + 5i \times (-5i) = 0$

$\implies x^2 – (0)x – 25i^2 = 0$

$\implies x^2 – 25(-1) = 0$

$\therefore x^2 + 25 = 0$

Question 2: Form the quadratic equation whose one root is:

Rule: If the coefficients of a quadratic equation are rational/real, then irrational and complex roots always occur in conjugate pairs.

(i) $\sqrt{3}i$

Solution:

Given one root, $\sqrt{3}i$

Then the other conjugate root must be $-\sqrt{3}i$

Sum of the roots:

$= \sqrt{3}i + (-\sqrt{3}i) $

$= 0$

Product of the roots:

$= (\sqrt{3}i)(-\sqrt{3}i) $

$ = -3i^2$ [Since $i^2 = -1$]

$= -3(-1) $

$ = 3$

∴ The required quadratic equation is:

$x^2 – (\text{Sum of roots})x + (\text{Product of roots}) = 0$

$\implies x^2 – (0)x + 3 = 0$

$\therefore x^2 + 3 = 0$

(i) $2$ less than the roots of $x^2 – 16x + 63 = 0$

Solution:

Let $\alpha$ and $\beta$ be the roots of $x^2 – 16x + 63 = 0$

$\therefore \alpha + \beta = 16 \quad \text{and} \quad \alpha\beta = 63$

Let new roots be $(\alpha – 2)$ and $(\beta – 2)$

Sum of new roots:

$(\alpha – 2) + (\beta – 2) $

$= (\alpha + \beta) – 4$

$= 16 – 4 $

$= 12$

Product of new roots:

$(\alpha – 2)(\beta – 2)$

$ = \alpha\beta – 2\alpha -2\beta + 4 $

$ = \alpha\beta – 2(\alpha + \beta) + 4 $

$ = 63 – 2(16) + 4 $

$ = 63 – 32 + 4 $

$= 35$

Therefore, the required equation is:

$x^2 – 12x + 35 = 0$

Question 4: Find the value of $k$ such that:

(i) one root of $2x^2 – 5x + k = 0$ is twice the other.

Solution:

Let one root be $\alpha$

Then the other roots will be $2\alpha$

$\text{Sum of roots: } \alpha + 2\alpha = \frac{-(-5)}{2} $

$\implies 3\alpha = \frac{5}{2} $

$\implies \alpha= \frac{5}{6}$

$\text{Product of roots: } \alpha \cdot (2\alpha)= \frac{k}{2} $

$ \implies 2\alpha^2= \frac{k}{2} $

$ \implies 4\alpha^2 = k$

$\implies k = 4 \left(\frac{5}{6}\right)^2 $

$\implies k = 4 \cdot \frac{25}{36} $

$\implies k= \frac{25}{9}$

Therefore, the required value of $k$ is $\frac{25}{9}$.

(ii) $(2k-5)x^2 – 4x – 15 = 0$ and $(3k-8)x^2 – 5x – 21 = 0$ have a common root.

Solution:

Let the common root be $\alpha$.

$\therefore (2k-5)\alpha^2 – 4\alpha – 15 = 0$ — (i)

$(3k-8)\alpha^2 – 5\alpha – 21 = 0$ — (ii)

By cross-multiplication, we have:

$\frac{\alpha^2}{(-4)(-21) – (-5)(-15)} = \frac{\alpha}{(-15)(3k-8) – (-21)(2k-5)} = \frac{1}{(2k-5)(-5) – (3k-8)(-4)}$

$\implies \frac{\alpha^2}{84-75} = \frac{\alpha}{-45k+120+42k-105} = \frac{1}{-10k+25+12k-32}$

$\implies \frac{\alpha^2}{9} = \frac{\alpha}{-3k+15} = \frac{1}{2k-7}$

From these, we can write:

$\frac{\alpha^2}{9} = \frac{1}{2k-7}$

$\implies \alpha^2 = \frac{9}{2k-7}$ — (iii)

and

$\frac{\alpha}{-3k+15} = \frac{1}{2k-7}$

$\implies \alpha = \frac{-3k+15}{2k-7}$ — (iv)

From (iii) and (iv), we get:

$\frac{9}{2k-7} = \left(\frac{-3k+15}{2k-7}\right)^2$

$\implies \frac{9}{2k-7} = \frac{9(5-k)^2}{(2k-7)^2}$

$\implies 1 = \frac{(5-k)^2}{2k-7}$

$\implies 2k-7 = 25 – 10k + k^2$

$\implies k^2 – 12k + 32 = 0$

$\implies k^2 – 8k – 4k + 32 = 0$

$\implies k(k-8) – 4(k-8) = 0$

$\implies (k-8)(k-4) = 0$

Either,

$k-8 = 0 \implies k = 8$

Or,

$k-4 = 0 \implies k = 4$

$\therefore$ The required values of $k$ are $8, 4$.

Question 5: Under what condition:

(i) $3x^2 + 4mx + 2 = 0$ and $2x^2 + 3x – 2 = 0$ will have a common root?

Solution:

Let $\alpha$ be the common root.

$\therefore 3\alpha^2 + 4m\alpha + 2 = 0$ — (i)

$2\alpha^2 + 3\alpha – 2 = 0$ — (ii)

From equation (ii):

$2\alpha^2 + 4\alpha – \alpha – 2 = 0$

$\implies 2\alpha(\alpha+2) – 1(\alpha+2) = 0$

$\implies (\alpha+2)(2\alpha-1) = 0$

$\implies \alpha = -2$ or $\alpha = \frac{1}{2}$

Case 1: When $\alpha = -2$, substituting this value in (i):

$3(-2)^2 + 4m(-2) + 2 = 0$

$\implies 12 – 8m + 2 = 0$

$\implies 8m = 14$

$\implies m = \frac{7}{4}$

Case 2: When $\alpha = \frac{1}{2}$, substituting this value in (i):

$3\left(\frac{1}{2}\right)^2 + 4m\left(\frac{1}{2}\right) + 2 = 0$

$\implies \frac{3}{4} + 2m + 2 = 0$

$\implies 2m = -\frac{11}{4}$

$\implies m = -\frac{11}{8}$

$\therefore$ The required conditions are $m = \frac{7}{4}$ or $m = -\frac{11}{8}$.

(ii) One root of $ax^2 + bx + c = 0$ will be $n$ times the other?

Solution:

Let one root be $\alpha$ and the other root be $n\alpha$.

Sum of the roots:

$\alpha + n\alpha = -\frac{b}{a}$

$\implies \alpha(1+n) = -\frac{b}{a}$

$\implies \alpha = \frac{-b}{a(1+n)}$ — (i)

Product of the roots:

$\alpha \times n\alpha = \frac{c}{a}$

$\implies n\alpha^2 = \frac{c}{a}$

$\implies \alpha^2 = \frac{c}{an}$ — (ii)

Substituting the value of $\alpha$ from (i) into (ii):

$\left(\frac{-b}{a(1+n)}\right)^2 = \frac{c}{an}$

$\implies \frac{b^2}{a^2(1+n)^2} = \frac{c}{an}$

$\implies \frac{b^2}{a(1+n)^2} = \frac{c}{n}$

$\therefore nb^2 = ac(1+n)^2$, which is the required condition.

(iii) One root of $x^2 – px + q = 0$ will be twice the other?

Solution:

Let one root be $\alpha$ and the other root be $2\alpha$.

Sum of the roots:

$\alpha + 2\alpha = p$

$\implies 3\alpha = p$

$\implies \alpha = \frac{p}{3}$ — (i)

Product of the roots:

$\alpha \times 2\alpha = q$

$\implies 2\alpha^2 = q$ — (ii)

Substituting the value of $\alpha$ from (i) into (ii):

$2\left(\frac{p}{3}\right)^2 = q$

$\implies 2\left(\frac{p^2}{9}\right) = q$

$\therefore 2p^2 = 9q$, which is the required condition.

(iv) The roots of $ax^2 + bx + c = 0$ will be in the ratio $m : n$?

Solution:

Let the roots be $m\alpha$ and $n\alpha$.

Sum of the roots:

$m\alpha + n\alpha = -\frac{b}{a}$

$\implies \alpha(m+n) = -\frac{b}{a}$

$\implies \alpha = \frac{-b}{a(m+n)}$ — (i)

Product of the roots:

$m\alpha \times n\alpha = \frac{c}{a}$

$\implies mn\alpha^2 = \frac{c}{a}$

Substituting the value of $\alpha$ from (i) into the product equation:

$mn\left(\frac{-b}{a(m+n)}\right)^2 = \frac{c}{a}$

$\implies mn \times \frac{b^2}{a^2(m+n)^2} = \frac{c}{a}$

$\implies \frac{mnb^2}{a(m+n)^2} = c$

$\therefore mnb^2 = ac(m+n)^2$, which is the required condition.

(v) $ax^2 + bx + c = 0$ and $px^2 + qx + r = 0$ will have a common root?

Solution:

Let $\alpha$ be the common root.

$\therefore a\alpha^2 + b\alpha + c = 0$ — (i)

$p\alpha^2 + q\alpha + r = 0$ — (ii)

By using the method of cross-multiplication, we get:

$\frac{\alpha^2}{br-cq} = \frac{\alpha}{cp-ar} = \frac{1}{aq-bp}$

$\implies \alpha^2 = \frac{br-cq}{aq-bp}$ — (iii)

and

$\alpha = \frac{cp-ar}{aq-bp}$

$\alpha^2 = \frac{(cp-ar)^2}{(aq-bp)^2}$ — (iv)

From (iii) and (iv), we get:

$\frac{(cp-ar)^2}{(aq-bp)^2} = \frac{br-cq}{aq-bp}$

$\implies \frac{(cp-ar)^2}{aq-bp} = br-cq$

$\therefore (cp-ar)^2 = (aq-bp)(br-cq)$, which is the required condition.

(vi) One root of $ax^2 + bx + c = 0$ is four times the other?

Solution:

Let one root be $\alpha$ and the other root be $4\alpha$.

Sum of the roots:

$\alpha + 4\alpha = -\frac{b}{a}$

$\implies 5\alpha = -\frac{b}{a}$

$\implies \alpha = -\frac{b}{5a}$ — (i)

Product of the roots:

$\alpha \times 4\alpha = \frac{c}{a}$

$\implies 4\alpha^2 = \frac{c}{a}$

Substituting the value of $\alpha$ from (i):

$4\left(-\frac{b}{5a}\right)^2 = \frac{c}{a}$

$\implies \frac{4b^2}{25a^2} = \frac{c}{a}$

$\therefore 4b^2 = 25ac$, which is the required condition.

(vii) The sum of the roots of $x^2 – mx + n = 0$ is $k$ times their difference?

Solution:

Let $\alpha$ and $\beta$ be the roots of the equation $x^2 – mx + n = 0$.

$\therefore \alpha + \beta = m \quad \text{and} \quad \alpha\beta = n$

According to the question:

$\alpha + \beta = k(\alpha – \beta)$

$(\alpha + \beta)^2 = k^2(\alpha – \beta)^2$ [Squaring both sides]

$\implies (\alpha + \beta)^2 = k^2[(\alpha + \beta)^2 – 4\alpha\beta]$

$\implies (\alpha + \beta)^2 = k^2(\alpha + \beta)^2 – 4k^2\alpha\beta$

$\implies 4k^2\alpha\beta = k^2(\alpha + \beta)^2 – (\alpha + \beta)^2$

$\implies 4k^2\alpha\beta = (\alpha + \beta)^2(k^2 – 1)$

Substituting $\alpha\beta = n$ and $\alpha + \beta = m$:

$\therefore 4k^2n = m^2(k^2 – 1)$, which is the required condition.

Question 6: If $\alpha$ and $\beta$ are the roots of $ax^2 + bx + c = 0$, express the value of the following symmetric functions in terms of the coefficients $a, b, c$:

(i) $\alpha^2 + \alpha\beta + \beta^2$

(ii) $(\alpha + 2\beta)(2\alpha + \beta)$

(iii) $\alpha^4 + \alpha^2\beta^2 + \beta^4$

(iv) $\frac{\alpha^2}{\beta} + \frac{\beta^2}{\alpha}$

Solution: 

Since $\alpha$ and $\beta$ are the roots of $ax^2 + bx + c = 0$:

$\therefore \alpha + \beta = -\frac{b}{a} \quad \text{and} \quad \alpha\beta = \frac{c}{a}$

(i) $\alpha^2 + \alpha\beta + \beta^2$

$\alpha^2 + \alpha\beta + \beta^2$

$= (\alpha^2 + \beta^2) + \alpha\beta$

$= [(\alpha + \beta)^2 – 2\alpha\beta] + \alpha\beta$

$= (\alpha + \beta)^2 – \alpha\beta$

$= \left(-\frac{b}{a}\right)^2 – \frac{c}{a}$

$= \frac{b^2}{a^2} – \frac{c}{a}$

$\therefore \frac{b^2 – ac}{a^2}$

(ii) $(\alpha + 2\beta)(2\alpha + \beta)$

Solution:

$(\alpha + 2\beta)(2\alpha + \beta) $

$= 2\alpha^2 + \alpha\beta + 4\alpha\beta + 2\beta^2$

$= 2(\alpha^2 + \beta^2) + 5\alpha\beta$

$= 2[(\alpha + \beta)^2 – 2\alpha\beta] + 5\alpha\beta$

$= 2(\alpha + \beta)^2 – 4\alpha\beta + 5\alpha\beta$

$= 2(\alpha + \beta)^2 + \alpha\beta$ 

$= 2\left(-\frac{b}{a}\right)^2 + \frac{c}{a}$

$= \frac{2b^2}{a^2} + \frac{c}{a}$

$\therefore \frac{2b^2 + ac}{a^2}$

(iii) $\alpha^4 + \alpha^2\beta^2 + \beta^4$

Solution:

$\alpha^4 + \alpha^2\beta^2 + \beta^4 $

$= (\alpha^4 + \beta^4) + \alpha^2\beta^2$

$= [(\alpha^2 + \beta^2)^2 – 2\alpha^2\beta^2] + \alpha^2\beta^2$

$= (\alpha^2 + \beta^2)^2 – \alpha^2\beta^2$

$= [(\alpha + \beta)^2 – 2\alpha\beta]^2 – (\alpha\beta)^2$

$= \left[ \left(-\frac{b}{a}\right)^2 – \frac{2c}{a} \right]^2 – \left(\frac{c}{a}\right)^2$

$= \left[ \frac{b^2}{a^2} – \frac{2c}{a} \right]^2 – \frac{c^2}{a^2}$

$= \left[ \frac{b^2 – 2ac}{a^2} \right]^2 – \frac{c^2}{a^2}$

$= \frac{(b^2 – 2ac)^2 – a^2c^2}{a^4}$

$= \frac{b^4 – 4ab^2c + 4a^2c^2 – a^2c^2}{a^4}$

$\therefore \frac{b^4 – 4ab^2c + 3a^2c^2}{a^4}$

(iv) $\frac{\alpha^2}{\beta} + \frac{\beta^2}{\alpha}$

Solution:

$\frac{\alpha^2}{\beta} + \frac{\beta^2}{\alpha} = \frac{\alpha^3 + \beta^3}{\alpha\beta}$

$= \frac{(\alpha + \beta)^3 – 3\alpha\beta(\alpha + \beta)}{\alpha\beta}$

$= \frac{\left(-\frac{b}{a}\right)^3 – 3\left(\frac{c}{a}\right)\left(-\frac{b}{a}\right)}{\frac{c}{a}}$

$= \frac{-\frac{b^3}{a^3} + \frac{3bc}{a^2}}{\frac{c}{a}}$

$= \left( \frac{-b^3 + 3abc}{a^3} \right) \times \frac{a}{c}$

$\therefore \frac{-b^3 + 3abc}{a^2c}$

Question 7: If $\alpha$ and $\beta$ are the roots of the equation $ax^2 + bx + c = 0$, find the quadratic equations having the following pairs of roots:

Recall: $\alpha + \beta = -\frac{b}{a}$ and $\alpha\beta = \frac{c}{a}$.

(i) $\frac{\alpha}{\beta}, \frac{\beta}{\alpha}$

Solution:

Sum of the roots:

$\frac{\alpha}{\beta} + \frac{\beta}{\alpha} $

$= \frac{\alpha^2 + \beta^2}{\alpha\beta}$

$= \frac{(\alpha + \beta)^2 – 2\alpha\beta}{\alpha\beta}$

$= \frac{\left(-\frac{b}{a}\right)^2 – \frac{2c}{a}}{\frac{c}{a}}$

$= \frac{b^2 – 2ac}{a^2} \times \frac{a}{c}$

$= \frac{b^2 – 2ac}{ac}$

Product of the roots:

$\frac{\alpha}{\beta} \times \frac{\beta}{\alpha} = 1$

The required quadratic equation is:

$x^2 – \left(\text{Sum of roots}\right)x + \text{Product of roots} = 0$

$\implies x^2 – \left(\frac{b^2 – 2ac}{ac}\right)x + 1 = 0$

$\therefore acx^2 – (b^2 – 2ac)x + ac = 0$

(ii) $\frac{1}{\alpha^2}, \frac{1}{\beta^2}$

Solution:

Sum of the roots:

$\frac{1}{\alpha^2} + \frac{1}{\beta^2} $

$= \frac{\alpha^2 + \beta^2}{\alpha^2\beta^2}$

$= \frac{(\alpha + \beta)^2 – 2\alpha\beta}{(\alpha\beta)^2}$

$= \frac{\left(-\frac{b}{a}\right)^2 – \frac{2c}{a}}{\left(\frac{c}{a}\right)^2}$

$= \frac{\frac{b^2 – 2ac}{a^2}}{\frac{c^2}{a^2}}$

$= \frac{b^2 – 2ac}{c^2}$

Product of the roots:

$\frac{1}{\alpha^2} \times \frac{1}{\beta^2} $

$= \frac{1}{(\alpha\beta)^2} $

$ = \frac{1}{\left(\frac{c}{a}\right)^2}$

$= \frac{a^2}{c^2}$

The required quadratic equation is:

$x^2 – \left(\frac{1}{\alpha^2} + \frac{1}{\beta^2}\right)x + \frac{1}{\alpha^2} \cdot \frac{1}{\beta^2} = 0$

$\implies x^2 – \left(\frac{b^2 – 2ac}{c^2}\right)x + \frac{a^2}{c^2} = 0$

$\therefore c^2x^2 – (b^2 – 2ac)x + a^2 = 0$

(iii) $\alpha^4, \beta^4$

Solution:

Sum of the roots:

$\alpha^4 + \beta^4 = (\alpha^2)^2 + (\beta^2)^2$

$= (\alpha^2 + \beta^2)^2 – 2\alpha^2\beta^2$

$= [(\alpha + \beta)^2 – 2\alpha\beta]^2 – 2(\alpha\beta)^2$

$= \left[ \left(-\frac{b}{a}\right)^2 – \frac{2c}{a} \right]^2 – 2\left(\frac{c}{a}\right)^2$

$= \left[ \frac{b^2 – 2ac}{a^2} \right]^2 – \frac{2c^2}{a^2}$

$= \frac{(b^2 – 2ac)^2 – 2a^2c^2}{a^4}$

$= \frac{b^4 – 4ab^2c + 2a^2c^2}{a^4}$

Product of the roots:

$\alpha^4 \times \beta^4 = (\alpha\beta)^4$

$= \left(\frac{c}{a}\right)^4 = \frac{c^4}{a^4}$

The required quadratic equation is:

$x^2 – (\alpha^4 + \beta^4)x + \alpha^4\beta^4 = 0$

$\implies x^2 – \left(\frac{b^4 – 4ab^2c + 2a^2c^2}{a^4}\right)x + \frac{c^4}{a^4} = 0$

$\therefore a^4x^2 – (b^4 – 4ab^2c + 2a^2c^2)x + c^4 = 0$

(iv) $\sqrt{\frac{\alpha}{\beta}}, \sqrt{\frac{\beta}{\alpha}}$

Solution:

Sum of the roots:

$\sqrt{\frac{\alpha}{\beta}} + \sqrt{\frac{\beta}{\alpha}} $

$ = \frac{\sqrt{\alpha}}{\sqrt{\beta}} + \frac{\sqrt{\beta}}{\sqrt{\alpha}}$

$= \frac{\alpha + \beta}{\sqrt{\alpha\beta}}$

$= \frac{-\frac{b}{a}}{\sqrt{\frac{c}{a}}}$

$= -\frac{b}{a} \times \sqrt{\frac{a}{c}}$

$= -\frac{b}{\sqrt{ac}}$

Product of the roots:

$\sqrt{\frac{\alpha}{\beta}} \times \sqrt{\frac{\beta}{\alpha}} = 1$

The required quadratic equation is:

$x^2 – \left(-\frac{b}{\sqrt{ac}}\right)x + 1 = 0$

$\therefore \sqrt{ac}x^2 + bx + \sqrt{ac} = 0$

(v) $\alpha^2 + \beta^2, \frac{1}{\alpha^2} + \frac{1}{\beta^2}$

Solution:

Sum of the roots:

$(\alpha^2 + \beta^2) + \left(\frac{1}{\alpha^2} + \frac{1}{\beta^2}\right) $

$= (\alpha^2 + \beta^2) + \left(\frac{\alpha^2 + \beta^2}{\alpha^2\beta^2}\right)$

$= (\alpha^2 + \beta^2)\left(1 + \frac{1}{\alpha^2\beta^2}\right)$

$= [(\alpha + \beta)^2 – 2\alpha\beta]\left(1 + \frac{1}{(\alpha\beta)^2}\right)$

$= \left( \frac{b^2 – 2ac}{a^2} \right)\left(1 + \frac{a^2}{c^2}\right)$

$= \left( \frac{b^2 – 2ac}{a^2} \right)\left( \frac{a^2 + c^2}{c^2} \right)$

$= \frac{(b^2 – 2ac)(a^2 + c^2)}{a^2c^2}$

Product of the roots:

$(\alpha^2 + \beta^2) \times \left(\frac{1}{\alpha^2} + \frac{1}{\beta^2}\right) $

$= (\alpha^2 + \beta^2) \times \left(\frac{\alpha^2 + \beta^2}{\alpha^2\beta^2}\right)$

$= \frac{(\alpha^2 + \beta^2)^2}{(\alpha\beta)^2}$

$= \frac{\left(\frac{b^2 – 2ac}{a^2}\right)^2}{\frac{c^2}{a^2}}$

$= \frac{(b^2 – 2ac)^2}{a^4} \times \frac{a^2}{c^2}$

$= \frac{(b^2 – 2ac)^2}{a^2c^2}$

The required quadratic equation is:

$x^2 – \left[ \frac{(b^2 – 2ac)(a^2 + c^2)}{a^2c^2} \right]x + \frac{(b^2 – 2ac)^2}{a^2c^2} = 0$

$\therefore a^2c^2x^2 – (b^2 – 2ac)(a^2 + c^2)x + (b^2 – 2ac)^2 = 0$

(vi) $\frac{1}{\alpha + \beta}, \frac{1}{\alpha} + \frac{1}{\beta}$

Solution:

Sum of the roots:

$\frac{1}{\alpha + \beta} + \left(\frac{1}{\alpha} + \frac{1}{\beta}\right) $

$= \frac{1}{\alpha + \beta} + \frac{\alpha + \beta}{\alpha\beta}$

$= \frac{1}{-\frac{b}{a}} + \frac{-\frac{b}{a}}{\frac{c}{a}}$

$= -\frac{a}{b} – \frac{b}{c}$

$= -\frac{ac + b^2}{bc}$

Product of the roots:

$\left(\frac{1}{\alpha + \beta}\right)\left(\frac{1}{\alpha} + \frac{1}{\beta}\right) $

$= \left(-\frac{a}{b}\right)\left(-\frac{b}{c}\right)$

$= \frac{a}{c}$

The required quadratic equation is:

$x^2 – \left[ -\frac{ac + b^2}{bc} \right]x + \frac{a}{c} = 0$

$\therefore bcx^2 + (ac + b^2)x + ab = 0$

(vii) $(\alpha – \beta)^2, (\alpha + \beta)^2$

Solution:

Sum of the roots:

$(\alpha – \beta)^2 + (\alpha + \beta)^2 $

$= 2(\alpha^2 + \beta^2)$

$= 2[(\alpha + \beta)^2 – 2\alpha\beta]$

$= 2\left[\frac{b^2}{a^2} – \frac{2c}{a}\right]$

$= \frac{2(b^2 – 2ac)}{a^2}$

Product of the roots:

$(\alpha – \beta)^2 \times (\alpha + \beta)^2 $

$= [(\alpha + \beta)^2 – 4\alpha\beta](\alpha + \beta)^2$

$= \left[ \left(-\frac{b}{a}\right)^2 – \frac{4c}{a} \right]\left(-\frac{b}{a}\right)^2$

$= \left( \frac{b^2 – 4ac}{a^2} \right)\left( \frac{b^2}{a^2} \right)$

$= \frac{b^2(b^2 – 4ac)}{a^4}$

The required quadratic equation is:

$x^2 – \left[ \frac{2(b^2 – 2ac)}{a^2} \right]x + \frac{b^2(b^2 – 4ac)}{a^4} = 0$

$\therefore a^4x^2 – 2a^2(b^2 – 2ac)x + b^2(b^2 – 4ac) = 0$

(viii) $\alpha + 2\beta, \beta + 2\alpha$

Solution:

Sum of the roots:

$(\alpha + 2\beta) + (\beta + 2\alpha) $

$= 3(\alpha + \beta)$

$= 3\left(-\frac{b}{a}\right) = -\frac{3b}{a}$

Product of the roots:

$(\alpha + 2\beta)(\beta + 2\alpha) $

$= \alpha\beta + 2\alpha^2 + 2\beta^2 + 4\alpha\beta$

$= 2(\alpha^2 + \beta^2) + 5\alpha\beta$

$= 2[(\alpha + \beta)^2 – 2\alpha\beta] + 5\alpha\beta$

$= 2(\alpha + \beta)^2 + \alpha\beta$

$= 2\left(-\frac{b}{a}\right)^2 + \frac{c}{a}$

$= \frac{2b^2 + ac}{a^2}$

The required quadratic equation is:

$x^2 – \left(-\frac{3b}{a}\right)x + \frac{2b^2 + ac}{a^2} = 0$

$\implies x^2 + \frac{3b}{a}x + \frac{2b^2 + ac}{a^2} = 0$

$\therefore a^2x^2 + 3abx + 2b^2 + ac = 0$

(ix) $\frac{\alpha^3}{\beta}, \frac{\beta^3}{\alpha}$

Solution:

Sum of the roots:

$\frac{\alpha^3}{\beta} + \frac{\beta^3}{\alpha} $

$ = \frac{\alpha^4 + \beta^4}{\alpha\beta}$

$= \frac{(\alpha^2 + \beta^2)^2 – 2\alpha^2\beta^2}{\alpha\beta}$

$= \frac{\{[(\alpha + \beta)^2 – 2\alpha\beta]\}^2 – 2(\alpha\beta)^2}{\alpha\beta}$

$= \frac{\left( \frac{b^2}{a^2}- \frac{2c}{a} \right)^2 – \frac{2c^2}{a^2}}{\frac{c}{a}}$

$= \frac{\left( \frac{b^2 – 2ac}{a^2} \right)^2 – \frac{2c^2}{a^2}}{\frac{c}{a}}$

$= \frac{\frac{b^4 – 4ab^2c + 4a^2c^2 – 2a^2c^2}{a^4}}{\frac{c}{a}}$

$= \frac{b^4 – 4ab^2c + 2a^2c^2}{a^3c}$

Product of the roots:

$\frac{\alpha^3}{\beta} \times \frac{\beta^3}{\alpha} = \alpha^2\beta^2 = (\alpha\beta)^2$

$= \frac{c^2}{a^2}$

The required quadratic equation is:

$x^2 – \left( \frac{b^4 – 4ab^2c + 2a^2c^2}{a^3c} \right)x + \frac{c^2}{a^2} = 0$

$\therefore a^3cx^2 – (b^4 – 4ab^2c + 2a^2c^2)x + ac^3 = 0$

(x) $\alpha^2 + \alpha\beta + \beta^2, \alpha^2 – \alpha\beta + \beta^2$

Solution:

Sum of the roots:

$(\alpha^2 + \alpha\beta + \beta^2) + (\alpha^2 – \alpha\beta + \beta^2) $

$= 2(\alpha^2 + \beta^2)$

$= 2\{(\alpha + \beta)^2 – 2\alpha\beta\}$

$= 2\left(\frac{b^2}{a^2} – \frac{2c}{a}\right)$

$= \frac{2b^2 – 4ac}{a^2}$

Product of the roots:

$(\alpha^2 + \beta^2 + \alpha\beta)(\alpha^2 + \beta^2 – \alpha\beta) $

$= (\alpha^2 + \beta^2)^2 – (\alpha\beta)^2$

$= \{(\alpha + \beta)^2 – 2\alpha\beta\}^2 – (\alpha\beta)^2$

$= \left( \frac{b^2 – 2ac}{a^2} \right)^2 – \frac{c^2}{a^2}$

$= \frac{b^4 – 4ab^2c + 4a^2c^2 – a^2c^2}{a^4}$

$= \frac{b^4 – 4ab^2c + 3a^2c^2}{a^4}$

The required quadratic equation is:

$x^2 – \left( \frac{2b^2 – 4ac}{a^2} \right)x + \frac{b^4 – 4ab^2c + 3a^2c^2}{a^4} = 0$

$\therefore a^4x^2 – a^2(2b^2 – 4ac)x + b^4 – 4ab^2c + 3a^2c^2 = 0$

Question 8: If $a^2 = 5a – 3$ and $b^2 = 5b – 3$ $(a \ne b)$, then find the quadratic equation whose roots are $\frac{a}{b}$ and $\frac{b}{a}$.

Solution:

Given equations:

$a^2 – 5a + 3 = 0$

$b^2 – 5b + 3 = 0$

This implies $a$ and $b$ are the roots of the quadratic equation $x^2 – 5x + 3 = 0$.

$\therefore \text{Sum of the roots} a + b = 5 $

$\text{Product of the roots} ab = 3$

Now,

Sum of the new roots:

$\frac{a}{b} + \frac{b}{a} = \frac{a^2 + b^2}{ab}$

$= \frac{(a+b)^2 – 2ab}{ab}$

$= \frac{5^2 – 2 \times 3}{3}$

$= \frac{19}{3}$

Product of the roots:

$\frac{a}{b} \times \frac{b}{a} = 1$

The required quadratic equation is:

$x^2 – \left( \frac{a}{b} + \frac{b}{a} \right)x + \left(\frac{a}{b} \times \frac{b}{a}\right) = 0$

$\implies x^2 – \frac{19}{3}x + 1 = 0$

$\therefore 3x^2 – 19x + 3 = 0$

Question 9: If $p$ and $q$ are the roots of $3x^2 + 6x + 2 = 0$, then find the quadratic equation having roots $-\frac{p^2}{q}$ and $-\frac{q^2}{p}$.

Solution:

Since $p$ and $q$ are the roots of $3x^2 + 6x + 2 = 0$:

$\therefore p+q = -\frac{6}{3} = -2$

$pq = \frac{2}{3}$

Now, we calculate for the new roots $-\frac{p^2}{q}$ and $-\frac{q^2}{p}$:

Sum of the roots:

$\left(-\frac{p^2}{q}\right) + \left(-\frac{q^2}{p}\right) = \frac{-p^3 – q^3}{pq}$

$= -\frac{(p^3 + q^3)}{pq}$

$= -\frac{[(p+q)^3 – 3pq(p+q)]}{pq}$

$= -\frac{\left[ (-2)^3 – 3\left(\frac{2}{3}\right)(-2) \right]}{\frac{2}{3}}$

$= -\frac{[-8 + 4]}{\frac{2}{3}}$

$= \frac{4 \times 3}{2} = 6$

Product of the roots:

$\left(-\frac{p^2}{q}\right) \times \left(-\frac{q^2}{p}\right) = pq = \frac{2}{3}$

The required quadratic equation is:

$x^2 – \left[ \left(-\frac{p^2}{q}\right) + \left(-\frac{q^2}{p}\right) \right]x + \left(-\frac{p^2}{q}\right)\left(-\frac{q^2}{p}\right) = 0$

$\implies x^2 – 6x + \frac{2}{3} = 0$

$\therefore 3x^2 – 18x + 2 = 0$

Question 10: If $4$ is a root of $x^2 + ax + 8 = 0$ and the roots of $x^2 + ax + b = 0$ are equal, then find the value of $b$.

Solution:

Since $4$ is a root of $x^2 + ax + 8 = 0$:

$\therefore 4^2 + a(4) + 8 = 0$

$\implies 16 + 4a + 8 = 0$

$\implies 4a = -24$

$\implies a = -6$

Now, substituting $a = -6$ into $x^2 + ax + b = 0$:

$x^2 – 6x + b = 0$

Since the roots of this equation are equal, its discriminant must be zero ($b^2 – 4ac = 0$):

$\therefore (-6)^2 – 4(1)(b) = 0$

$\implies 36 – 4b = 0$

$\implies 4b = 36$

$\therefore b = 9$

Question 11: If $\alpha$ be one root of $4x^2 + 2x – 1 = 0$, then show that the other root is $4\alpha^3 – 3\alpha$.

Solution:

Let $\alpha$ and $\beta$ be the roots of $4x^2 + 2x – 1 = 0$.

We need to prove that $\beta = 4\alpha^3 – 3\alpha$.

Now, $4x^2 + 2x – 1 = 0$:

Here, $a=4, b=2, c=-1$.

$\therefore x = \frac{-2 \pm \sqrt{2^2 – 4(4)(-1)}}{2(4)}$

$= \frac{-2 \pm \sqrt{4 + 16}}{8}$

$= \frac{-2 \pm \sqrt{20}}{8}$

$= \frac{-2 \pm 2\sqrt{5}}{8}$

$= \frac{-1 \pm \sqrt{5}}{4}$

Let $\alpha = \frac{-1 + \sqrt{5}}{4}$, then the other root is $\beta = \frac{-1 – \sqrt{5}}{4}$.

Now,

$4\alpha^3 – 3\alpha = 4\left(\frac{-1+\sqrt{5}}{4}\right)^3 – 3\left(\frac{-1+\sqrt{5}}{4}\right)$

$= 4\left( \frac{(-1+\sqrt{5})^3}{64} \right) – 3\left(\frac{-1+\sqrt{5}}{4}\right)$

$= 4\left( \frac{(-1)^3 + 3(-1)^2(\sqrt{5}) + 3(-1)(\sqrt{5})^2 + (\sqrt{5})^3}{64} \right) – 3\left(\frac{-1+\sqrt{5}}{4}\right)$

$= 4\left( \frac{-1 + 3\sqrt{5} – 15 + 5\sqrt{5}}{64} \right) – 3\left(\frac{-1+\sqrt{5}}{4}\right)$

$= \frac{-16 + 8\sqrt{5}}{16} – \frac{-3 + 3\sqrt{5}}{4}$

$= \frac{-2 + \sqrt{5}}{2} – \frac{-3 + 3\sqrt{5}}{4}$

$= \frac{2(-2 + \sqrt{5}) – (-3 + 3\sqrt{5})}{4}$

$= \frac{-4 + 2\sqrt{5} + 3 – 3\sqrt{5}}{4}$

$= \frac{-1 – \sqrt{5}}{4}$

$= \beta$

$\therefore$ Showed.

Question 12: If the difference of the two roots of $x^2 + px + q = 0$ is $1$, then show that $p^2 + 4q^2 = (1 + 2q)^2$.

Solution:

Let $\alpha$ and $\alpha-1$ be the two roots of the equation $x^2 + px + q = 0$.

Sum of the roots:

$\alpha + (\alpha – 1) = -p$

$\implies 2\alpha – 1 = -p$

$\implies 2\alpha = 1 – p$

$\implies \alpha = \frac{1-p}{2}$ — (i)

Product of the roots:

$\alpha(\alpha – 1) = q$

Substituting the value of $\alpha$ from (i):

$\left(\frac{1-p}{2}\right)\left(\frac{1-p}{2} – 1\right) = q$

$\implies \left(\frac{1-p}{2}\right)\left(\frac{1-p-2}{2}\right) = q$

$\implies \frac{1-p}{2} \times \frac{-(1+p)}{2} = q$

$\implies -\frac{(1-p^2)}{4} = q$

$\implies -(1-p^2) = 4q$

$\implies p^2 – 1 = 4q$

$\implies p^2 = 4q + 1$ — (ii)

Now, substituting $p^2 = 4q + 1$ into the LHS of our proof:

$\text{LHS} = p^2 + 4q^2$

$= 4q + 1 + 4q^2$

$= 1 + 4q + 4q^2$

$= (1 + 2q)^2 = \text{RHS}$

$\therefore$ Hence proved.

Question 13: If $ax^2 + bx + c = 0$ and $bx^2 + cx + a = 0$ have a common root, then prove that $a + b + c = 0$ or $a = b = c$.

Solution:

Let $\alpha$ be the common root of both equations.

$\therefore a\alpha^2 + b\alpha + c = 0$ — (i)

$b\alpha^2 + c\alpha + a = 0$ — (ii)

By using cross-multiplication, we get:

$\frac{\alpha^2}{ab – c^2} = \frac{\alpha}{bc – a^2} = \frac{1}{ac – b^2}$

$\implies \alpha^2 = \frac{ab – c^2}{ac – b^2}$ — (iii)

and

$\alpha = \frac{bc – a^2}{ac – b^2} $

$\implies \alpha^2 = \frac{(bc – a^2)^2}{(ac – b^2)^2}$ — (iv)

From (iii) and (iv), we get:

$\frac{ab – c^2}{ac – b^2} = \frac{(bc – a^2)^2}{(ac – b^2)^2}$

$\implies ab – c^2 = \frac{(bc – a^2)^2}{ac – b^2}$

$\implies (ab – c^2)(ac – b^2) = (bc – a^2)^2$

$\implies a^2bc – ab^3 – ac^3 + b^2c^2 = b^2c^2 – 2a^2bc + a^4$

$\implies a^4 + ab^3 + ac^3 – 3a^2bc = 0$

$\implies a(a^3 + b^3 + c^3 – 3abc) = 0$

$\implies a^3 + b^3 + c^3 – 3abc = 0$

$\implies (a + b + c)(a^2 + b^2 + c^2 – ab – bc – ca) = 0$

$\implies \frac{1}{2}(a + b + c)[(a – b)^2 + (b – c)^2 + (c – a)^2] = 0$

Either,

$a + b + c = 0$

Or,

$[(a – b)^2 + (b – c)^2 + (c – a)^2] = 0$

Since the sum of three perfect squares is zero, each individual term must be zero:

$(a-b)^2 = 0$

$\implies(a-b) = 0$

$ \implies a = b$

Similarly,

$(b-c)^2 = 0 \implies b = c$

$(c-a)^2 = 0 \implies c = a$

$\implies a = b = c$

$\therefore$ Proved.

Question 14: If the two roots of $ax^2 + bx + a = 0$ are equal, then show that $\frac{a^2+b^2}{a^2-b^2} = -\frac{5}{3}$.

Solution:

Since the roots of $ax^2 + bx + a = 0$ are equal, its discriminant must be zero:

$b^2 – 4a^2 = 0$ [Here constant term $c$ is actually $a$] 

$\implies b^2 = 4a^2$ — (i)

Now, LHS:

$\frac{a^2 + b^2}{a^2 – b^2}$

$ = \frac{a^2 + 4a^2}{a^2 – 4a^2}$ [From eq(i)]

$= \frac{5a^2}{-3a^2}$

$= -\frac{5}{3}$

$\therefore$ Showed.

Question 15: Solve:

(i) $x^4 – 13x^2 + 36 = 0$

Solution:

Let $x^2 = y$.

$\implies y^2 – 13y + 36 = 0$

$\implies y^2 – 9y – 4y + 36 = 0$

$\implies y(y – 9) – 4(y – 9) = 0$

$\implies (y – 9)(y – 4) = 0$

Either,

$y – 9 = 0 \implies y = 9 $

$\implies x^2 = 9 \implies x = \pm 3$

Or,

$y – 4 = 0 \implies y = 4 $

$\implies x^2 = 4 \implies x = \pm 2$

$\therefore$ Required roots: $3, -3, 2, -2$.

(ii) $x^4 – 3x^2 + 2 = 0$

Solution:

Let $x^2 = y$.

$\implies y^2 – 3y + 2 = 0$

$\implies y^2 – 2y – y + 2 = 0$

$\implies y(y-2) – 1(y-2) = 0$

$\implies (y-2)(y-1) = 0$

Either,

$y – 2 = 0 \implies y = 2 $

$\implies x^2 = 2 \implies x = \pm \sqrt{2}$

Or,

$y – 1 = 0 \implies y = 1 $

$\implies x^2 = 1 \implies x = \pm 1$

$\dots$ Required roots: $\sqrt{2}, -\sqrt{2}, 1, -1$.

(iii) $(x^2 – 3x)^2 – 5(x^2 – 3x) + 6 = 0$

Solution:

Let $x^2 – 3x = y$.

$\implies y^2 – 5y + 6 = 0$

$\implies y^2 – 3y – 2y + 6 = 0$

$\implies y(y-3) – 2(y-3) = 0$

$\implies (y-3)(y-2) = 0$

Either,

$y – 3 = 0 \implies x^2 – 3x – 3 = 0$

Using the quadratic formula:

$x = \frac{-(-3) \pm \sqrt{(-3)^2 – 4(1)(-3)}}{2(1)}$

$x = \frac{3 \pm \sqrt{9 + 12}}{2} = \frac{3 \pm \sqrt{21}}{2}$

Or,

$y – 2 = 0 \implies x^2 – 3x – 2 = 0$

Using the quadratic formula:

$x = \frac{-(-3) \pm \sqrt{(-3)^2 – 4(1)(-2)}}{2(1)}$

$x = \frac{3 \pm \sqrt{9 + 8}}{2} = \frac{3 \pm \sqrt{17}}{2}$

$\therefore$ Required roots: $\frac{3\pm\sqrt{21}}{2}, \frac{3\pm\sqrt{17}}{2}$.

(iv) $(x^2 + 2x – 3)^2 – 3(x^2 + 2x – 1) + 8 = 0$

Solution:

Let $x^2 + 2x – 1 = a$.

$\implies x^2 + 2x – 3 = a – 2$

Substituting these into the equation:

$(a – 2)^2 – 3a + 8 = 0$

$\implies a^2 – 4a + 4 – 3a + 8 = 0$

$\implies a^2 – 7a + 12 = 0$

$\implies a^2 – 4a – 3a + 12 = 0$

$\implies a(a – 4) – 3(a – 4) = 0$

$\implies (a-4)(a-3) = 0$

Either,

$a – 4 = 0 \implies a = 4$

$\implies x^2 + 2x – 1 = 4$

$\implies x^2 + 2x – 5 = 0$

Using the quadratic formula:

$x = \frac{-2 \pm \sqrt{2^2 – 4(1)(-5)}}{2}$

$\implies x = \frac{-2 \pm \sqrt{24}}{2} = \frac{-2 \pm 2\sqrt{6}}{2} = -1 \pm \sqrt{6}$

Or,

$a – 3 = 0 \implies a = 3$

$\implies x^2 + 2x – 1 = 3$

$\implies x^2 + 2x – 4 = 0$

Using the quadratic formula:

$x = \frac{-2 \pm \sqrt{2^2 – 4(1)(-4)}}{2}$

$\implies x = \frac{-2 \pm \sqrt{20}}{2} = \frac{-2 \pm 2\sqrt{5}}{2} = -1 \pm \sqrt{5}$

$\therefore$ Required roots: $-1 \pm \sqrt{6}, -1 \pm \sqrt{5}$.

(v) $x^2 – 5x + 10 = 5\sqrt{x^2 – 5x + 4}$

Solution:

Let $x^2 – 5x + 4 = a $

$\implies x^2 – 5x + 10 = a + 6$.

$\implies a + 6 = 5\sqrt{a}$

Squaring both sides:

$(a+6)^2 = (5\sqrt{a})^2$

$\implies a^2 + 12a + 36 = 25a$

$\implies a^2 – 13a + 36 = 0$

$\implies a^2 – 9a – 4a + 36 = 0$

$\implies a(a-9) – 4(a-9) = 0$

$\implies (a-9)(a-4) = 0$

Either,

$a – 9 = 0 \implies x^2 – 5x + 4 – 9 = 0 $

$\implies x^2 – 5x – 5 = 0$

Using the quadratic formula:

$x = \frac{-(-5) \pm \sqrt{(-5)^2 – 4(1)(-5)}}{2}$

$\implies x = \frac{5 \pm \sqrt{45}}{2} = \frac{5 \pm 3\sqrt{5}}{2}$

Or,

$a – 4 = 0 \implies x^2 – 5x + 4 – 4 = 0 $

$\implies x(x-5) = 0$

$\implies x = 0$ or $x = 5$

$\therefore$ Required roots: $0, 5, \frac{5 \pm 3\sqrt{5}}{2}$.

(vi) $\sqrt{x^2 + 5x – 2} + \sqrt{x^2 + 5x – 5} = 3$

Solution:

Let $x^2 + 5x – 2 = u$.

$\implies (x^2 + 5x – 2) – 3 = u – 3$

$\implies x^2 + 5x – 5 = u – 3$

$\therefore\sqrt{u} + \sqrt{u-3} = 3$

$\implies \sqrt{u} = 3 – \sqrt{u-3}$

Squaring both sides:

$u = (3 – \sqrt{u-3})^2$

$\implies u = 9 + (u-3) – 6\sqrt{u-3}$

$\implies u = u + 6 – 6\sqrt{u-3}$

$\implies -6 = -6\sqrt{u-3}$

$\implies 1 = \sqrt{u-3}$

Squaring both sides:

$1 = u-3$

$\implies u = 4$

Substitute $u = x^2 + 5x – 2$:

$x^2 + 5x – 2 = 4$

$\implies x^2 + 5x – 6 = 0$

$\implies x^2 + 6x – x – 6 = 0$

$\implies x(x+6) – 1(x+6) = 0$

$\implies (x+6)(x-1) = 0$

Either,

$x + 6 = 0 \implies x = -6$

Or,

$x – 1 = 0 \implies x = 1$

$\therefore$ Required roots: $-6, 1$.

(vii) $\sqrt{\frac{x}{1-x}} + \sqrt{\frac{1-x}{x}} = \frac{13}{6}$

Solution:

Let $\sqrt{\frac{x}{1-x}} = u $

$\implies \sqrt{\frac{1-x}{x}} = \frac{1}{u}$.

$\therefore u + \frac{1}{u} = \frac{13}{6}$

$\implies \frac{u^2 + 1}{u} = \frac{13}{6}$

$\implies 6(u^2 +1) = 13u $

$\implies 6u^2 + 6 = 13u $

$\implies 6u^2 – 13u + 6 = 0$

$\implies 6u^2 – 9u – 4u + 6 = 0$

$\implies 3u(2u – 3) – 2(2u – 3) = 0$

$\implies (2u – 3)(3u – 2) = 0$

Either,

$2u – 3 = 0 \implies u = \frac{3}{2}$

$\implies \sqrt{\frac{x}{1-x}} = \frac{3}{2} \implies \frac{x}{1-x} = \frac{9}{4}$

$\implies 4x = 9 – 9x \implies 13x = 9 $

$\implies x = \frac{9}{13}$

Or,

$3u – 2 = 0 \implies u = \frac{2}{3}$

$\implies \sqrt{\frac{x}{1-x}} = \frac{2}{3} \implies \frac{x}{1-x} = \frac{4}{9}$

$\implies 9x = 4 – 4x \implies 13x = 4$

$\implies x = \frac{4}{13}$

$\therefore$ Required roots: $\frac{9}{13}, \frac{4}{13}$.

(viii) $\left(x^2 + \frac{1}{x^2}\right) – 5\left(x + \frac{1}{x}\right) = 4$

Solution:

We now, $\left(x + \frac{1}{x}\right)^2 = x^2 + 2 + \frac{1}{x^2}$

$\implies x^2 + \frac{1}{x^2} = \left(x + \frac{1}{x}\right)^2 – 2$

$\therefore \left[\left(x + \frac{1}{x}\right)^2 – 2\right] – 5\left(x + \frac{1}{x}\right) = 4$

$\implies \left(x + \frac{1}{x}\right)^2 – 5\left(x + \frac{1}{x}\right) – 6 = 0$

Let $x + \frac{1}{x} = a$.

$\implies a^2 – 5a – 6 = 0$

$\implies a^2 – (6 – 1) a – 6 = 0$

$\implies a^2 – 6a + a – 6 = 0$

$\implies a(a-6) + 1(a-6) = 0$

$\implies (a-6)(a+1) = 0$

Either,

$a – 6 = 0 \implies x + \frac{1}{x} = 6$

$\implies \frac{x^2 + 1}{x} = 6 $

$\implies x^2 – 6x + 1 = 0$

Using the quadratic formula:

$x = \frac{-(-6) \pm \sqrt{(-6)^2 – 4(1)(1)}}{2}$

$\implies x = \frac{6 \pm \sqrt{32}}{2}$

$\implies x = \frac{6 \pm 4\sqrt{2}}{2} = 3 \pm 2\sqrt{2}$

Or,

$a + 1 = 0 $

$\implies x + \frac{1}{x} = -1$

$\implies x^2 + x + 1 = 0$

Using the quadratic formula:

$x = \frac{-1 \pm \sqrt{1^2 – 4(1)(1)}}{2}$

$\implies x = \frac{-1 \pm \sqrt{-3}}{2} = \frac{-1 \pm \sqrt{3}i}{2}$

$\therefore$ Required roots: $3 \pm 2\sqrt{2}, \frac{-1 \pm \sqrt{3}i}{2}$.

(ix) $(x-1)(x-2)(x-3)(x-4) = 120$

Solution:

$\implies [(x-1)(x-4)][(x-2)(x-3)] = 120$

$\implies (x^2 – 5x + 4)(x^2 – 5x + 6) = 120$

Let $x^2 – 5x = a$.

$\implies (a+4)(a+6) = 120$

$\implies a^2 + 10a + 24 – 120 = 0$

$\implies a^2 + 10a – 96 = 0$

$\implies a^2 + (16 – 6)a – 96 = 0$

$\implies a^2 + 16a – 6a – 96 = 0$

$\implies a(a+16) – 6(a+16) = 0$

$\implies (a+16)(a-6) = 0$

Either,

$a + 16 = 0 $

$\implies x^2 – 5x + 16 = 0$

Using the quadratic formula:

$x = \frac{-(-5) \pm \sqrt{(-5)^2 – 4(1)(16)}}{2}$

$\implies x = \frac{5 \pm \sqrt{25 – 64}}{2} $

$ = \frac{5 \pm \sqrt{-39}}{2} $

$\therefore x = \frac{5 \pm \sqrt{39}i}{2}$

Or,

$a – 6 = 0 $

$ \implies x^2 – 5x – 6 = 0$

$\implies x^2 – 6x + x – 6 = 0$

$\implies x(x-6) + 1(x-6) = 0$

$\implies (x-6)(x+1) = 0$

$\implies x = 6$ or $x = -1$

$\therefore$ Required roots: $6, -1, \frac{5 \pm \sqrt{39}i}{2}$.

(x) $(x+1)(x+3)(x+5)(x+7) = 20$

Solution:

$\implies [(x+1)(x+7)][(x+3)(x+5)] = 20$

$\implies (x^2 + 8x + 7)(x^2 + 8x + 15) = 20$

Let $x^2 + 8x = a$.

$\implies (a+7)(a+15) = 20$

$\implies a^2 + 22a + 105 – 20 = 0$

$\implies a^2 + 22a + 85 = 0$

$\implies a^2 + (17 + 5)a + 85 = 0$

$\implies a^2 + 17a + 5a + 85 = 0$

$\implies a(a+17) + 5(a+17) = 0$

$\implies (a+17)(a+5) = 0$

Either,

$a + 17 = 0 $

$\implies x^2 + 8x + 17 = 0$

Using the quadratic formula:

$x = \frac{-8 \pm \sqrt{8^2 – 4(1)(17)}}{2}$

$\implies x = \frac{-8 \pm \sqrt{64 – 68}}{2} $

$= \frac{-8 \pm \sqrt{-4}}{2} $

$= \frac{-8 \pm 2i}{2} $

$ = -4 \pm i$

Or,

$a + 5 = 0 $

$\implies x^2 + 8x + 5 = 0$

Using the quadratic formula:

$x = \frac{-8 \pm \sqrt{8^2 – 4(1)(5)}}{2}$

$\implies x = \frac{-8 \pm \sqrt{64 – 20}}{2} $

$ = \frac{-8 \pm \sqrt{44}}{2} $

$ = \frac{-8 \pm 2\sqrt{11}}{2}$

$= -4 \pm \sqrt{11}$

$\therefore$ Required roots: $-4 \pm i, -4 \pm \sqrt{11}$.

(xi) $\frac{1}{x-2} + \frac{1}{x+5} = \frac{1}{x-6}$

Solution:

$\frac{1}{x-2} + \frac{1}{x+5} = \frac{1}{x-6}$

$\implies \frac{(x+5) + (x-2)}{(x-2)(x+5)} = \frac{1}{x-6}$

$\implies \frac{2x+3}{x^2+3x-10} = \frac{1}{x-6}$

$\implies (2x+3)(x-6) = x^2 + 3x – 10$

$\implies 2x^2 – 12x + 3x – 18 = x^2 + 3x – 10$

$\implies x^2 – 12x – 8 = 0$

Using the quadratic formula:

$x = \frac{-(-12) \pm \sqrt{(-12)^2 – 4(1)(-8)}}{2}$

$\implies x = \frac{12 \pm \sqrt{144 + 32}}{2}$

$\implies x = \frac{12 \pm \sqrt{176}}{2}$

$\implies x = \frac{12 \pm 4\sqrt{11}}{2} = 6 \pm 2\sqrt{11}$

$\therefore$ Required roots: $6 \pm 2\sqrt{11}$.

(xii) $\frac{5}{x^2+6x+8} = \frac{1}{x^2+6x+5} + \frac{4}{x^2+6x+9}$

Solution:

Let $x^2 + 6x = a$.

$\implies \frac{5}{a+8} = \frac{1}{a+5} + \frac{4}{a+9}$

$\implies \frac{4}{a+8} + \frac{1}{a+8} = \frac{1}{a+5} + \frac{4}{a+9}$

$\implies \frac{4}{a+8} – \frac{4}{a+9} = \frac{1}{a+5} – \frac{1}{a+8}$

$\implies 4\left[\frac{1}{a+8} – \frac{1}{a+9}\right] = \frac{1}{a+5} – \frac{1}{a+8}$

$\implies 4\left[\frac{a+9-a-8}{(a+8)(a+9)}\right] = \frac{a+8-a-5}{(a+5)(a+8)}$

$\implies \frac{4}{(a+8)(a+9)} = \frac{3}{(a+5)(a+8)}$

$\implies 4(a+8)(a+5) = 3(a+8)(a+9)$

$\implies 4(a+8)(a+5) – 3(a+8)(a+9) = 0$

$\implies (a+8)[4(a+5) – 3(a+9)] = 0$

$\implies (a+8)(4a + 20 – 3a – 27) = 0$

$\implies (a+8)(a – 7) = 0$

Either,

$a + 8 = 0$

$\implies x^2 + 6x + 8 = 0$

$\implies x^2 + 4x + 2x + 8 = 0$

$\implies x(x+4) + 2(x+4) = 0$

$\implies (x+4)(x+2) = 0$

$\implies x = -4$ or $x = -2$

Or,

$a – 7 = 0$

$\implies x^2 + 6x – 7 = 0$

$\implies x^2 + 7x – x – 7 = 0$

$\implies x(x+7) – 1(x+7) = 0$

$\implies (x+7)(x-1) = 0$

$\implies x = -7$ or $x = 1$

$\therefore$ Required roots: $-7, -4, -2, 1$

(xiii) $\frac{x^2-5}{2x-3} – \frac{3x^2}{6x+1} = \frac{3}{2}$

Solution:

Taking LCM on the Left Hand Side:

$\implies \frac{(x^2-5)(6x+1) – 3x^2(2x-3)}{(2x-3)(6x+1)} = \frac{3}{2}$

$\implies \frac{(6x^3 + x^2 – 30x – 5) – (6x^3 – 9x^2)}{12x^2 + 2x – 18x – 3} = \frac{3}{2}$

$\implies \frac{10x^2 – 30x – 5}{12x^2 – 16x – 3} = \frac{3}{2}$

$\implies 2(10x^2 – 30x – 5) = 3(12x^2 – 16x – 3)$

$\implies 20x^2 – 60x – 10 = 36x^2 – 48x – 9$

$\implies 16x^2 + 12x + 1 = 0$

Using the quadratic formula:

$x = \frac{-12 \pm \sqrt{12^2 – 4(16)(1)}}{2(16)}$

$\implies x = \frac{-12 \pm \sqrt{144 – 64}}{32}$

$\implies x = \frac{-12 \pm \sqrt{80}}{32}$

$\implies x = \frac{-12 \pm 4\sqrt{5}}{32}$

$\implies x = \frac{4(-3 \pm \sqrt{5})}{32}$

$\therefore x = \frac{-3 \pm \sqrt{5}}{8}$

(xiv) $\sqrt{2x-1} + \sqrt{3x-2} = \sqrt{4x-3} + \sqrt{5x-4}$

Solution:

The given equation is:

$\sqrt{2x-1} + \sqrt{3x-2} = \sqrt{4x-3} + \sqrt{5x-4}$

Rearranging the terms to group them strategically:

$\implies \sqrt{2x-1} – \sqrt{5x-4} = \sqrt{4x-3} – \sqrt{3x-2}$

Squaring both sides, we get:

$\implies (\sqrt{2x-1} – \sqrt{5x-4})^2 = (\sqrt{4x-3} – \sqrt{3x-2})^2$

$\implies 2x – 1 – 2\sqrt{2x-1}\cdot\sqrt{5x-4} + 5x – 4 = 4x – 3 – 2\sqrt{4x-3}\cdot\sqrt{3x-2} + 3x – 2$

$\implies 7x – 5 – 2\sqrt{2x-1}\cdot\sqrt{5x-4} = 7x – 5 – 2\sqrt{4x-3}\cdot\sqrt{3x-2}$

Canceling $7x – 5$ from both sides:

$\implies -2\sqrt{2x-1}\cdot\sqrt{5x-4} = -2\sqrt{4x-3}\cdot\sqrt{3x-2}$

Dividing both sides by $-2$:

$\implies \sqrt{2x-1}\cdot\sqrt{5x-4} = \sqrt{4x-3}\cdot\sqrt{3x-2}$

Squaring both sides again:

$\implies (\sqrt{2x-1}\cdot\sqrt{5x-4})^2 = (\sqrt{4x-3}\cdot\sqrt{3x-2})^2$

$\implies (2x – 1)(5x – 4) = (4x – 3)(3x – 2)$

$\implies 10x^2 – 8x – 5x + 4 = 12x^2 – 8x – 9x + 6$

$\implies 10x^2 – 13x + 4 = 12x^2 – 17x + 6$

Rearranging all terms to one side:

$\implies 2x^2 – 4x + 2 = 0$

$\implies 2(x^2 – 2x + 1) = 0$

$\implies x^2 – 2x + 1 = 0$

$\implies (x – 1)^2 = 0$

$\implies (x – 1)(x – 1) = 0$

Either,

$x – 1 = 0 \implies x = 1$

Or,

$x – 1 = 0 \implies x = 1$

$\therefore x = 1, 1$

Required roots: $1, 1$

(xv) $\sqrt{x^2-4} + \sqrt{x^2+5x+6} = \sqrt{3x^2+13x+14}$

Solution:

By factoring the expressions under the square roots, we get:

$\implies \sqrt{(x-2)(x+2)} + \sqrt{(x+2)(x+3)} = \sqrt{(x+2)(3x+7)}$

$\implies \sqrt{x+2} \left( \sqrt{x-2} + \sqrt{x+3} – \sqrt{3x+7} \right) = 0$

Either,

$\sqrt{x+2} = 0 \implies x + 2 = 0 \implies x = -2$

Or,

$\sqrt{x-2} + \sqrt{x+3} – \sqrt{3x+7} = 0$

$\implies \sqrt{x-2} + \sqrt{x+3} = \sqrt{3x+7}$

Squaring both sides:

$\implies (x-2) + (x+3) + 2\sqrt{(x-2)(x+3)} = 3x+7$

$\implies 2x + 1 + 2\sqrt{x^2+x-6} = 3x+7$

$\implies 2\sqrt{x^2+x-6} = x+6$

Squaring both sides again:

$\implies 4(x^2+x-6) = (x+6)^2$

$\implies 4x^2 + 4x – 24 = x^2 + 12x + 36$

$\implies 3x^2 – 8x – 60 = 0$

$\implies 3x^2 – 18x + 10x – 60 = 0$

$\implies 3x(x-6) + 10(x-6) = 0$

$\implies (3x+10)(x-6) = 0$

$\implies x = 6$ or $x = -\frac{10}{3}$

• For $x = -\frac{10}{3}$, the terms under $\sqrt{x-2}$ and $\sqrt{x^2-4}$ become negative, so it is not acceptable.

$\therefore$ Required roots: $-2, 6$.

(xvi) $3^{x+3} + 3^x – 3^{2x+1} = 9$

Solution:

$\implies 3^3 \cdot 3^x + 3^x – 3 \cdot (3^x)^2 = 9$

$\implies 27 \cdot 3^x + 3^x – 3 \cdot (3^x)^2 = 9$

$\implies 28 \cdot 3^x – 3(3^x)^2 = 9$

Let $3^x = y$.

$\implies 28y – 3y^2 = 9$

$\implies 3y^2 – 28y + 9 = 0$

$\implies 3y^2 – 27y – y + 9 = 0$

$\implies 3y(y-9) – 1(y-9) = 0$

$\implies (3y-1)(y-9) = 0$

Either,

$3y – 1 = 0 $

$ \implies y = \frac{1}{3} $

$\implies 3^x = 3^{-1} $

$\implies x = -1$

Or,

$y – 9 = 0 $

$\implies y = 9 $

$\implies 3^x = 3^2 $

$ \implies x = 2$

$\therefore$ Required roots: $-1, 2$.

(xvi) $4^x – 3 \cdot 2^{x+2} + 32 = 0$

Solution:

$\implies (2^2)^x – 3 \cdot 2^x \cdot 2^2 + 32 = 0$

$\implies (2^x)^2 – 12 \cdot 2^x + 32 = 0$

Let $2^x = y$.

$\implies y^2 – 12y + 32 = 0$

$\implies y^2 – 8y – 4y + 32 = 0$

$\implies y(y-8) – 4(y-8) = 0$

$\implies (y-8)(y-4) = 0$

Either,

$y – 8 = 0 \implies y = 8 $

$\implies 2^x = 2^3 $

$\implies x = 3$

Or,

$y – 4 = 0 \implies y = 4 $

$\implies 2^x = 2^2 $

$\implies x = 2$

$\therefore$ Required roots: $2, 3$.

(xviii) $x^{\frac{2}{3}} – x^{\frac{1}{3}} – 2 = 0$

Solution:

Let $x^{\frac{1}{3}} = y$.

$\implies y^2 – y – 2 = 0$

$\implies y^2 – 2y + y – 2 = 0$

$\implies y(y-2) + 1(y-2) = 0$

$\implies (y-2)(y+1) = 0$

Either,

$y – 2 = 0 \implies y = 2 $

$\implies x^{\frac{1}{3}} $

$= 2 \implies x = 2^3 $

$\implies x = 8$

Or,

$y + 1 = 0 \implies y = -1 $

$\implies x^{\frac{1}{3}} = -1 $

$\implies x = (-1)^3 $

$\implies x = -1$

$\therefore$ Required roots: $-1, 8$.

(xix) $x^{-4} – 10x^{-2} + 9 = 0$

Solution:

Let $x^{-2} = y$.

$\implies y^2 – 10y + 9 = 0$

$\implies y^2 – 9y – y + 9 = 0$

$\implies y(y-9) – 1(y-9) = 0$

$\implies (y-9)(y-1) = 0$

Either,

$y – 9 = 0 \implies y = 9 $

$\implies x^{-2} = 9 $

$\implies \frac{1}{x^2} = 9$

$\implies x^2 = \frac{1}{9}$

$\implies x = \pm \frac{1}{3}$

Or,

$y – 1 = 0 $

$\implies y = 1 $

$\implies x^{-2} = 1 $

$\implies \frac{1}{x^2} = 1$

$\implies x^2 = 1 $

$\implies x = \pm 1$

$\therefore$ Required roots: $\pm 1, \pm \frac{1}{3}$.

(xx) $3^{2x} + 9 = 10\left(\frac{1}{3}\right)^{-x}$

Solution:

Since $\left(\frac{1}{3}\right)^{-x} = (3^{-1})^{-x} = 3^x$, the equation becomes:

$\implies (3^x)^2 + 9 = 10 \cdot 3^x$

$\implies (3^x)^2 – 10 \cdot 3^x + 9 = 0$

Let $3^x = y$.

$\implies y^2 – 10y + 9 = 0$

$\implies y^2 – 9y – y + 9 = 0$

$\implies y(y-9) – 1(y-9) = 0$

$\implies (y-9)(y-1) = 0$

Either,

$y – 9 = 0 $

$\implies y = 9 $

$\implies 3^x = 3^2$

$\implies x = 2$

Or,

$y – 1 = 0 $

$\implies y = 1 $

$\implies 3^x = 3^0 $

$\implies x = 0$

$\therefore$ Required roots: $0, 2$.