SEBA Class 10 Maths Exercise 4.4 Solutions: Quadratic Equations

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SEBA Class 10 Maths Exercise 4.1 Solutions
SEBA Class 10 Maths Exercise 4.2 Solutions
SEBA Class 10 Maths Exercise 4.3 Solutions

1. Find the nature of the roots of the following quadratic equations. If the real roots exist, find them:

(i) $2x^{2}-3x+5=0$

Solⁿ : 

Given : $2x^{2}-3x+5=0$

Here $a=2$, $b=-3$, and $c=5$.

∴ The discriminant, $b^{2}-4ac$
$=(-3)^{2}-(4\times2\times5)$
$=9-40$
$=-31<0$
$b^{2}-4ac<0$

Hence, the given equation has no real roots.

(ii) $3x^{2}-4\sqrt{3}x+4=0$

Solⁿ : 

Given : $3x^{2}-4\sqrt{3}x+4=0$

Here, $a=3$, $b=-4\sqrt{3}$, and $c=4$.

∴ The discriminant, $b^{2}-4ac$
$=(-4\sqrt{3})^{2}-(4\times3\times4)$
$=(16\times3)-48$
$=48-48$
$=0$
$b^{2}-4ac=0$

Hence, the roots are real and equal.

Now, using the quadratic formula:

$x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$, we get
$=\frac{-b}{2a}\pm0$
$=\frac{-(-4\sqrt{3})}{2\times3}$
$=\frac{4\sqrt{3}}{2\times3}$
$=\frac{2\sqrt{3}}{3}$
$=\frac{2\sqrt{3}}{\sqrt{3}\times\sqrt{3}}$
$=\frac{2}{\sqrt{3}}$

Therefore, the equal roots of the given equation are $\frac{2}{\sqrt{3}}$ and $\frac{2}{\sqrt{3}}$.

(iii) $2x^{2}-6x+3=0$

Soln :

Given : $2x^{2}-6x+3=0$

Here, $a=2$, $b=-6$, and $c=3$.

∴ The discriminant, $b^{2}-4ac$

$=(-6)^{2}-(4\times2\times3)$

$=36-24$

$=12>0$

$b^{2}-4ac>0$

Hence, the roots are real and distinct.

Now, using the quadratic formula:

$x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$, we get
$=\frac{-(-6)\pm\sqrt{12}}{2\times2}$
$=\frac{6\pm2\sqrt{3}}{4}$
$=\frac{2(3\pm\sqrt{3})}{4}$
$=\frac{3\pm\sqrt{3}}{2}$

Therefore, the roots of the given equation are $\frac{3+\sqrt{3}}{2}$ and $\frac{3-\sqrt{3}}{2}$.

(iv) $9x^{2}-6x+1=0$

Soln :

Given : $9x^{2}-6x+1=0$

Here, $a=9$, $b=-6$, and $c=1$.

∴ The discriminant, $b^{2}-4ac$

$=(-6)^{2}-(4\times9\times1)$

$=36-36$

$=0$

$b^{2}-4ac=0$

Hence, the roots are real and equal.

Now, using the quadratic formula:
$x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$, we get
$=\frac{-(-6)\pm0}{2\times9}$
$=\frac{6}{18}$ $=\frac{1}{3}$

Therefore, the equal roots of the given equation are $\frac{1}{3}$ and $\frac{1}{3}$.

(v) $3x^{2}-5x+12=0$

Soln :

Given : $3x^{2}-5x+12=0$

Here, $a=3$, $b=-5$, and $c=12$.

∴ The discriminant, $b^{2}-4ac$

$=(-5)^{2}-(4\times3\times12)$

$=25-144$

$=-119<0$

$b^{2}-4ac<0$

Hence, the given equation has no real roots.

(vi) $x^{2}+x+1=0$

Soln :

Given : $x^{2}+x+1=0$

Here, $a=1$, $b=1$, and $c=1$.

∴ The discriminant, $b^{2}-4ac$

$=(1)^{2}-(4\times1\times1)$

$=1-4$

$=-3<0$

$b^{2}-4ac<0$

Hence, the given equation has no real roots.

(vii) $x^{2}-2\sqrt{3}x-9=0$

Soln :

Given : $x^{2}-2\sqrt{3}x-9=0$

Here, $a=1$, $b=-2\sqrt{3}$, and $c=-9$.

∴ The discriminant, $b^{2}-4ac$

$=(-2\sqrt{3})^{2}-(4\times1\times(-9))$

$=12+36$

$=48>0$

$b^{2}-4ac>0$

Hence, the roots are real and distinct.

Now, using the quadratic formula:

$x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$, we get
$=\frac{-(-2\sqrt{3})\pm\sqrt{48}}{2\times1}$
$=\frac{2\sqrt{3}\pm4\sqrt{3}}{2}$
$=\sqrt{3}\pm2\sqrt{3}$

Either, $x=\sqrt{3}+2\sqrt{3}$
$=3\sqrt{3}$

or, $x=\sqrt{3}-2\sqrt{3}$
$=-\sqrt{3}$

Therefore, the roots of the given equation are $3\sqrt{3}$ and $-\sqrt{3}$.

(viii) $2x^{2}-3\sqrt{3}x+3=0$

Soln :

Given : $2x^{2}-3\sqrt{3}x+3=0$

Here, $a=2$, $b=-3\sqrt{3}$, and $c=3$.

∴ The discriminant, $b^{2}-4ac$

$=(-3\sqrt{3})^{2}-(4\times2\times3)$

$=27-24$

$=3>0$

$b^{2}-4ac>0$

Hence, the roots are real and distinct.

Now, using the quadratic formula:

$x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$, we get

$=\frac{-(-3\sqrt{3})\pm\sqrt{3}}{2\times2}$

$=\frac{3\sqrt{3}\pm\sqrt{3}}{4}$

Either, $x=\frac{3\sqrt{3}+\sqrt{3}}{4}$
$=\frac{4\sqrt{3}}{4}=\sqrt{3}$

or, $x=\frac{3\sqrt{3}-\sqrt{3}}{4}$
$=\frac{2\sqrt{3}}{4}$
$=\frac{\sqrt{3}}{2}$

Therefore, the roots of the given equation are $\sqrt{3}$ and $\frac{\sqrt{3}}{2}$.

2. Find the values of k for each of the following quadratic equations, so that they have two equal roots.

(i) $2x^{2}+kx+3=0$

Soln :

Given : $2x^{2}+kx+3=0$

which is of the form $ax^{2}+bx+c=0$

Here, $a=2$, $b=k$, and $c=3$.

∴ The discriminant, $b^{2}-4ac$

$=k^{2}-(4\times2\times3)$

$=k^{2}-24$

The given equation will have two equal real roots if $b^{2}-4ac=0$
⇒ $k^{2}-24=0$
⇒ $k^{2}=24$
⇒ $k=\pm\sqrt{24}$
⇒ $k=\pm2\sqrt{6}$

Therefore, the required value of $k$ is $\pm2\sqrt{6}$.

(ii) $kx(x-2)+6=0$

Soln :

Given : $kx(x-2)+6=0$

⇒ $kx^{2}-2kx+6=0$

which is of the form $ax^{2}+bx+c=0$

Here, $a=k$, $b=-2k$, and $c=6$.

∴ The discriminant, $b^{2}-4ac$

$=(-2k)^{2}-(4\times k\times6)$

$=4k^{2}-24k$

The given equation will have two equal real roots if $b^{2}-4ac=0$
⇒ $4k^{2}-24k=0$
⇒ $4k(k-6)=0$
⇒ $k(k-6)=0$

Either, $k=0$

or $k-6=0 $
$\Rightarrow k=6$

But $k=0$ is not admissible, as for $k=0$, the coefficient $a$ becomes zero and the given quadratic equation reduces to $6=0$, which is not possible.

Therefore, the required value of $k$ is $6$.

(iii) $x^{2}-(k+4)x+2k+5=0$

Soln :

Given : $x^{2}-(k+4)x+2k+5=0$

which is of the form $ax^{2}+bx+c=0$

Here, $a=1$, $b=-(k+4)$, and $c=2k+5$.

∴ The discriminant, $b^{2}-4ac$

$=\{-(k+4)\}^{2}-4\times1\times(2k+5)$

$=k^{2}+8k+16-8k-20$

$=k^{2}-4$

The given equation will have two equal real roots if $b^{2}-4ac=0$

⇒ $k^{2}-4=0$

⇒ $k^{2}=4$

⇒ $k=\pm2$

Therefore, the required values of $k$ are $\pm2$.

(iv) $2x^{2}+8x-k^{3}=0$

Soln :

Given : $2x^{2}+8x-k^{3}=0$

which is of the form $ax^{2}+bx+c=0$

Here, $a=2$, $b=8$, and $c=-k^{3}$.

∴ The discriminant, $b^{2}-4ac$

$=8^{2}-4\times2\times(-k^{3})$

$=64+8k^{3}$

The given equation will have two equal real roots if $b^{2}-4ac=0$
⇒ $64+8k^{3}=0$
⇒ $8k^{3}=-64$
⇒ $k^{3}=-8$
⇒ $k^{3}=(-2)^{3}$
⇒ $k=-2$

Therefore, the required value of $k$ is $-2$.

(v) $(k-3)x^{2}+6x+9=0$

Soln :

Given : $(k-3)x^{2}+6x+9=0$

which is of the form $ax^{2}+bx+c=0$

Here, $a=k-3$, $b=6$, and $c=9$.

∴ The discriminant, $b^{2}-4ac$

$=6^{2}-4\times(k-3)\times9$

$=36-36(k-3)$

$=36-36k+108$

$=-36k+144$

The given equation will have two equal real roots if $b^{2}-4ac=0$
⇒ $-36k+144=0$
⇒ $-36k=-144$
⇒ $k=4$

Therefore, the required value of $k$ is $4$.

(vi) $(k-12)x^{2}+2(k-12)x+2=0$

Soln :

Given : $(k-12)x^{2}+2(k-12)x+2=0$

which is of the form $ax^{2}+bx+c=0$

Here, $a=k-12$, $b=2(k-12)$, and $c=2$.

∴ The discriminant, $b^{2}-4ac$

$=\{2(k-12)\}^{2}-4\times(k-12)\times2$

$=4(k-12)^{2}-8(k-12)$

The given equation will have two equal real roots if $b^{2}-4ac=0$
⇒ $4(k-12)^{2}-8(k-12)=0$
⇒ $(k-12)^{2}-2(k-12)=0$
⇒ $(k-12)(k-12-2)=0$
⇒ $(k-12)(k-14)=0$

Either, $k-12=0$
$\Rightarrow k=12$

or, $k-14=0 $
$\Rightarrow k=14$

But $k=12$ is not admissible, as for $k=12$, the coefficient $a=k-12$ becomes zero and the equation is no longer quadratic.

Therefore, the required value of $k$ is $14$.

3. Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is $800\text{ m}^{2}$? If so, find its length and breadth.

Soln :

Let the breadth of the rectangular mango grove be $x\text{ m}$.

∴ Length of the grove $=2x\text{ m}$

Area of the rectangular mango grove $=2x\times x\text{ m}^{2}$

$=2x^{2}\text{ m}^{2}$

According to the question,

$2x^{2}=800$

⇒ $x^{2}=\frac{800}{2}$

⇒ $x^{2}=400$

⇒ $x^{2}-400=0$

⇒ $x^{2}-0x-400=0$

Here, $a=1$, $b=0$, and $c=-400$.

∴ The discriminant, $b^{2}-4ac$

$=0^{2}-4\times1\times(-400)$

$=1600>0$

$b^{2}-4ac>0$

Therefore, the above equation has two distinct real roots, and it is possible to design the given rectangular mango grove.

Now, $x^{2}=400$

⇒ $x=\pm20$

Either, $x=20$ or $x=-20$

Since $x=-20$ is not admissible as distance/length cannot be negative, $x=20$

Therefore, the breadth of the grove is $20\text{ m}$ and

its length is $2x=2\times20\text{ m}=40\text{ m}$.

4. Is the following situation possible? If so, determine their present ages. The sum of the ages of two friends is $20\text{ years}$. Four years ago, the product of their ages in years was $48$.

Soln :

Let $A$ and $B$ be two friends, and the present age of $A$ be $x\text{ years}$.

Present age of $B=(20-x)\text{ years}$.

Four years ago, $A\text{‘s age}=(x-4)\text{ years}$

$B\text{‘s age}=(20-x-4)\text{ years}=(16-x)\text{ years}$

According to the question, $(x-4)(16-x)=48$
⇒ $x(16-x)-4(16-x)=48$
⇒ $16x-x^{2}-64+4x=48$
⇒ $-x^{2}+20x-64=48$
⇒ $x^{2}-20x+64+48=0$
⇒ $x^{2}-20x+112=0$

Here, $a=1$, $b=-20$, and $c=112$.

∴ The discriminant, $b^{2}-4ac$

$=(-20)^{2}-4\times1\times112$

$=400-448$

$=-48<0$

$b^{2}-4ac<0$

Therefore, the above quadratic equation does not have real roots.

Hence, the given situation is not possible.

5. Is it possible to design a rectangular park of perimeter $80\text{ m}$ and area $400\text{ m}^{2}$? If so, find its length and breadth.

Soln :

Let the length of the park be $x\text{ m}$.

∴ Breadth of the park $=\frac{400}{x}\text{ m}$

Perimeter of the park $=2(\text{length}+\text{breadth})=80\text{ m}$

According to the question,

$2(x+\frac{400}{x})=80$

⇒ $x+\frac{400}{x}=\frac{80}{2}$

⇒ $\frac{x^{2}+400}{x}=40$

⇒ $x^{2}+400=40x$

⇒ $x^{2}-40x+400=0$

Here, $a=1$, $b=-40$, and $c=400$.

∴ The discriminant, $b^{2}-4ac$

$=(-40)^{2}-4\times1\times400$

$=1600-1600$

$=0$

$b^{2}-4ac=0$

Therefore, the above quadratic equation has two equal real roots, and it is possible to design the given park.

Now, $x^{2}-40x+400=0$

⇒ $x^{2}-20x-20x+400=0$

⇒ $x(x-20)-20(x-20)=0$

⇒ $(x-20)(x-20)=0$

⇒ $x=20, 20$

Length of the park $(x)=20\text{ m}$

Breadth of the park $(\frac{400}{x})=\frac{400}{20}=20\text{ m}$

Therefore, the length and breadth of the rectangular park are $20\text{ m}$ and $20\text{ m}$ respectively.

6. The equation $9x^{2}-6x-2=0$ has

(a) no real root

(b) two equal roots

(c) two distinct roots

(d) more than two real roots

Soln :

Given : $9x^{2}-6x-2=0$

Here, $a=9$, $b=-6$, and $c=-2$.

∴ The discriminant, $b^{2}-4ac$

$=(-6)^{2}-4\times9\times(-2)$

$=36+72$

$=108>0$

Since $b^{2}-4ac>0$, the equation has two distinct real roots.

Correct Option : (c)

7. Match column I with the nature of the roots given in column II.

Soln :

For P: $x^{2}-2x+3=0 $

$ \Rightarrow D=(-2)^{2}-4(1)(3) $

$ =4-12$

$ =-8<0 $

$\Rightarrow$ Roots are not real (iii)

For Q: $3x^{2}-2x+\frac{1}{3}=0 $

$ \Rightarrow D=(-2)^{2}-4(3)(\frac{1}{3})$

$=4-4=0 $

$\Rightarrow$ Roots are real and equal (iv)

For R: $x^{2}-4x+3=0 $

$\Rightarrow D=(-4)^{2}-4(1)(3)$

$=16-12=4>0$ (perfect square) $

$\Rightarrow$ Roots are real and unequal/rational (ii)

For S: $2x^{2}-x-4=0 $
$\Rightarrow D=(-1)^{2}-4(2)(-4)$
$=1+32=33>0$ (not a perfect square)
$\Rightarrow$ Roots are real and irrational (i)

Hence, $P\rightarrow(iii), Q\rightarrow(iv), R\rightarrow(ii), S\rightarrow(i)$

Correct Option : (b)

8. Assertion (A): If one root of the quadratic equation $(k-1)x^{2}-10x+3=0$ is reciprocal to the other root then $k=4$

Reason (R): One root of the quadratic equation $ax^{2}+bx+c=0$, $(a\ne0)$ reciprocal to other only when $a=c$

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of the Assertion (A)
(c) Assertion (A) is true but Reason (R) is false.
(d) Assetion (A) is false but Reason (R) is true.

Soln :

Let the roots of $ax^{2}+bx+c=0$ be $\alpha$ and $\frac{1}{\alpha}$.

Product of roots $=\alpha\times\frac{1}{\alpha}=1$

Also, product of roots $=\frac{c}{a}$

∴ $\frac{c}{a}=1 \Rightarrow a=c$.

So, Reason (R) is true.

For Assertion (A): Given equation $(k-1)x^{2}-10x+3=0$, where $a=k-1$ and $c=3$.

Since roots are reciprocal to each other, $a=c$
⇒ $k-1=3$
⇒ $k=4$

So, Assertion (A) is true and Reason (R) is the correct explanation of Assertion (A).

Correct Option : (a)