ML Aggarwal Solution Class 10 Chapter 5 Quadratic Equations in One Variable Exercise 5.1

 Exercise 5.1

Question 1

Check whether the following are quadratic equations:

(i) $\sqrt{3} x^{2}-2 x+\frac{3}{5}=0$
(ii)(2x + 1) (3x – 2) = 6(x + 1) (x – 2)
(iii) $(x-3)^{3}+5=x^{3}+7 x^{2}-1$
(iv) $x-\frac{3}{x}=2, x \neq 0$
(v) $x+\frac{2}{x}=x^{2}, x \neq 0$
(vi) $x^{2}+\frac{1}{x^{2}}=3, x \neq 0$
Sol :
(i) 
$\sqrt{3} x^{2}-2 x+\frac{3}{5}=0$

It is a quadratic equation as it is power of 2.


(ii)

(2x + 1) (3x – 2) = 6(x + 1) (x – 2)

6x² – 4x + 3x – 2 = 6(x² – 2x + x – 2)

6x² – x – 2 = 6x² – 12x + 6x – 12

12x-6x-x=-12+2

5x=-10

It is not a quadratic equation.


(iii) 

$(x-3)^{3}+5=x^{3}+7 x^{2}-1$

$x^{3}-3 x^{2} \times 3+3 x \times 9-27+5=x^{3}+7 x^{2}-1$

$-9 x^{2}+27 x-22-7 x^{2}+1=0$

$-16 x^{2}+27 x-21=0$

$16 x^{2}-27 x+21=0$

It is a quadratic equation


(iv)

$x-\frac{3}{x}=2, x \neq 0$

$x^{2}-3=2 x \Rightarrow x^{2}-2 x-3=0$

It is a quadratic equation


(v)

$x+\frac{2}{x}=x^{2}, x \neq 0$

$x^{2}+2=x^{3} \Rightarrow x^{3}-x^{2}-2=0$

It is not a quadratic equation


(vi)

$x^{2}+\frac{1}{x^{2}}=3, x \neq 0$
$x^{4}+6=3 x^{2}$
$x^{4}-3 x^{2}+6=0$
It is not a quadratic equation

Question 2

In each of the following, determine whether the given numbers are roots of the given equations or not;

(i) x² – x + 1 = 0; 1, – 1

(ii) x² – 5x + 6 = 0; 2, – 3

(iii) 3x² – 13x – 10 = 0; 5,$\frac{-2}{3}$

(iv) 6x² – x – 2 = 0; $\frac{-1}{2}, \frac{2}{3}$

Sol :

(i)

 x² – x + 1 = 0; 1, -1

Where x = 1, then

(1)² – 1 + 1 = 1 – 1 + 1 = 1 ≠ 0

∴ x = 1 does not satisfy it

and $(-1)^{2}-(-1)+1=0$
$1+1+1 \Rightarrow 3 \neq 0$

∴x=-1, does not satisfy it

∴x=1,-1 are not roots of the equation


(ii)

$x^{2}-5 x+6=0 ; 2,-3$

When x=2, then

$(2)^{2}-5 \times 2+6$

=4-10+6=10-10=0

∴x=2 is its root

Where x=-3, then

$(-3)^{2}-5(-3)+6$

$=9+15+6=30 \neq 0$

∴x=-3 is not its solution

∴2 is root of the equation by -3 is not a root.


(iii)

$3 x^{2}-13 x-10=0 ; 5, \frac{-2}{3}$

x=5

$3(5)^{2}-13 \times 5-10=75-65-10$

=75-75=0

∴x=5 is it root

If $x=\frac{-2}{3},$ then

$3\left(\frac{-2}{3}\right)^{2}-13 \times \frac{-2}{3}-10$

$=\frac{3 \times 4}{9}+\frac{26}{3}-10$

$=\frac{4}{3}+\frac{26}{3}-10=\frac{30}{3}-10=10-10=0$

$\therefore x=\frac{-2}{3}$ is also its root.

Hence both $5, \frac{-2}{3}$ are its roots.


(iv)

$6 x^{2}-x-2=0 ; \frac{-1}{2}, \frac{2}{3}$

If $x=\frac{-1}{2},$ then

$=6\left(\frac{-1}{2}\right)^{2}-\left(\frac{-1}{2}\right)-2$

$=6 \times \frac{1}{4}+\frac{1}{2}-2=\frac{3}{2}+\frac{1}{2}-2$

$=\frac{4}{2}-2=0$

$\therefore x=\frac{-1}{2}$ is its root

If $x=\frac{2}{3},$ then

$=6 \times \frac{4}{9}-\frac{2}{3}-2=\frac{8}{3}-\frac{2}{3}-2$

$=\frac{6}{3}-2=0$

$\therefore x=\frac{2}{3}$ is also its root.

Hence $\frac{-1}{2}, \frac{2}{3}$ are both its root.


Question 3

In each of the following, determine whether the given numbers are solutions of the given equation or not:

(i) x² – 3√3x + 6 = 0; √3, – 2√3

(ii) x² – √2x – 4 = 0, x = – √2, 2√2

Sol :

(i) x² – 3√3x + 6 = 0; √3, -2√3

(a) Substituting the value of x = √3

L.H.S. $=x^{2}-3 \sqrt{3} x+6$

$=(\sqrt{3})^{2}-3 \sqrt{3} \times \sqrt{3}+6=3-9+6=0$=R.H.S

∴$x=\sqrt{3}$ is its solution


(b) $x=-2 \sqrt{3}$

Substituting $x=-2 \sqrt{3}$

L.H.S. $=x^{2}-3 \sqrt{3} x+6$

$=(-2 \sqrt{3})^{2}-3 \sqrt{3}(-2 \sqrt{3})+6$

=12+18+6=36≠0

∵$x=-2 \sqrt{3}$ is not its solution


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

(a) $x=-\sqrt{2}$

Substituting $x=-\sqrt{2}$

$\mathrm{L.H.S.}=x^{2}-\sqrt{2} x-4$

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

=2+2-4=0=R.H.S
∴$x=-\sqrt{2}$ is its solution


(b) $x=-2 \sqrt{2}$

Substituting $x=-2 \sqrt{2}$

L.H.S. $=x^{2}-\sqrt{2} x-4$

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

=8-4-4=8-8=0=R.H.S

∴$x=-2 \sqrt{2}$ is its solution


Question 4

(i) If $-\frac{1}{2}$ is a solution of the equation 3x² + 2kx – 3 = 0, find the value of k.

(ii) If $\frac{2}{3}$ is a solution of the equation 7x² + kx – 3 = 0, find the value of k.

Sol : 

$x=-\frac{1}{2}$ is a solution of the

3x² + 2kx – 3 = 0,

Substituting the value of x in the given equation

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

$3 \times \frac{1}{4}-k-3=0$

$\frac{3}{4}-k-3=0$

$\Rightarrow k=\frac{3}{4}-3=-\frac{9}{4}$

Hence $k=-\frac{9}{4}$


(ii) $7 x^{2}+k x-3=0, x=\frac{2}{3}$

$\because x=\frac{2}{3}$ is its solution

$\therefore 7\left(\frac{2}{3}\right)^{2}+k\left(\frac{2}{3}\right)-3=0$

$\Rightarrow 7 \times \frac{4}{9}+\frac{2}{3} k-3=0$

$\Rightarrow \frac{28}{9}-3+\frac{2}{3} k=0$

$\Rightarrow \frac{2}{3} k=3-\frac{28}{9}$

$\Rightarrow \frac{2}{3} k=\frac{27-28}{9} \Rightarrow \frac{2}{3} k=\frac{-1}{9}$

$\Rightarrow k=\frac{-1}{9} \times \frac{3}{2}=\frac{-1}{6}$

Hence $k=\frac{-1}{6}$


Question 5

(i) If √2 is a root of the equation kx² + √2 – 4 = 0, find the value of k.

(ii) If a is a root of the equation x² – (a + b)x + k = 0, find the value of k.

Sol :

(i) kx² + √2 – 4 = 0, x = √2

x = √2 is its solution

$\therefore k(\sqrt{2})^{2}+\sqrt{2} \times \sqrt{2}-4=0$
$\Rightarrow 2 k+2-4=0$
$\Rightarrow 2 k-2=0$
$\Rightarrow 2 k=2$
$\Rightarrow k=\frac{2}{2}=1$
$\therefore k=1$

(ii) $x^{2}-x(a+b)+k=0, x=a$
$\because x=a$ is its solution
$\therefore(a)^{2}-a(a+b)+k=0$
$\Rightarrow a^{2}-a^{2}-a b+k=0 \Rightarrow-a b+k=0$
$\therefore k=a b$


Question 6

If $\frac{2}{3}$ and – 3 are the roots of the equation px² + 7x + q = 0, find the values of p and q.

Sol :
$\frac{2}{3}$ and – 3 are the roots of the equation px² + 7x + q = 0

Substituting the value of $x=\frac{2}{3}$ and – 3 respectively, we get

⇒$p\left(\frac{2}{3}\right)^{2}+7\left(\frac{2}{3}\right)+q=0$

⇒$\frac{4}{9} p+\frac{14}{3}+q=0$

⇒4 p+42+9 q=0

⇒4p+9q=-42...(i)

and $p(-3)^{2}+7(-3)+q=0$

⇒9p-21+q=0

⇒q=21-9p...(ii)

Substituting the value of q in (i)

⇒4 p+9(21-9 p)=-42

⇒4 p+189-81 p=-42

⇒-77 p=-42-189=-231

⇒$p=\frac{-231}{-77}=3$

∴q=21-9×3

=21-27=-6

∴p=3, q=-6

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