SEBA Class 10 Advanced Maths Chapter 4.1 Solutions: Quadratic Equation – Part-2
Get free SEBA Class 10 Advanced Maths Chapter 4.1 Solutions: Quadratic Equation. This is Part 2 of the solutions for Chapter 2 Advanced Mathematics Books from SEBA. This article provides 100% reliable and accurate solutions for all the remaining questions of Exercise 4.1 from the SEBA Class 10 Advanced Maths textbook for the upcoming HSLC Examination.
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SEBA Class 10 Advanced Maths Chapter 4.1 Solutions: Quadratic Equation – Part-1
Q11: If $\alpha$ be one root of $4x^2 + 2x – 1 = 0$, then show that the other root is $4\alpha^3 – 3\alpha$.
Soln:
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.
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Q12: If the difference of the two roots of $x^2 + px + q = 0$ is $1$, then show that $p^2 + 4q^2 = (1 + 2q)^2$.
Soln:
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.
Q13: If $ax^2 + bx + c = 0$ and $bx^2 + cx + a = 0$ have a common root, then prove that $a + b + c = 0$ or $a = b = c$.
Soln:
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.
Q14: If the two roots of $ax^2 + bx + a = 0$ are equal, then show that $\frac{a^2+b^2}{a^2-b^2} = -\frac{5}{3}$.
Soln:
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.
Q15: Solve:
(i) $x^4 – 13x^2 + 36 = 0$
Soln:
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) $x^4 – 3x^2 + 2 = 0$
Soln:
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, we get:
$\implies \sqrt{2x-1} – \sqrt{5x-4} = \sqrt{4x-3} – \sqrt{3x-2}$
Squaring both sides, we get
$\implies (\sqrt{2x-1} – \sqrt{5x-4})^2 = (\sqrt{4x-3} – \sqrt{3x-2})^2$
$\implies 2x – 1 – 2\sqrt{2x-1}\cdot\sqrt{5x-4} + 5x – 4 = 4x – 3 – 2\sqrt{4x-3}\cdot\sqrt{3x-2} + 3x – 2$
$\implies 7x – 5 – 2\sqrt{2x-1}\cdot\sqrt{5x-4} = 7x – 5 – 2\sqrt{4x-3}\cdot\sqrt{3x-2}$
$\implies -2\sqrt{2x-1}\cdot\sqrt{5x-4} = -2\sqrt{4x-3}\cdot\sqrt{3x-2}$ [Canceling $7x – 5$ from both sides]
$\implies \sqrt{2x-1}\cdot\sqrt{5x-4} = \sqrt{4x-3}\cdot\sqrt{3x-2}$ [dividing both sides by $-2$]
$\implies (\sqrt{2x-1}\cdot\sqrt{5x-4})^2 = (\sqrt{4x-3}\cdot\sqrt{3x-2})^2$ [Squaring both sides again]
$\implies (2x – 1)(5x – 4) = (4x – 3)(3x – 2)$
$\implies 10x^2 – 8x – 5x + 4 = 12x^2 – 8x – 9x + 6$
$\implies 10x^2 – 13x + 4 = 12x^2 – 17x + 6$
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: $0, 2$.