Question

Match the quadratic equations given in Column I with their roots in Column II and choose the correct option from the codes given below.

Column I		Column II
A	$\sqrt{2} x^2+x+\sqrt{2}=0$	1	$x=\frac{\sqrt{2} \pm i \sqrt{34}}{2 \sqrt{3}}$
B	$\sqrt{3} x^2-\sqrt{2} x+3 \sqrt{3}=0$	2	$x=\frac{-1 \pm i \sqrt{7}}{2 \sqrt{2}}$
C	$x^2+x+\frac{1}{\sqrt{2}}=0$	3	$\frac{-1 \pm i \sqrt{7}}{2 \sqrt{2}}$
D	$x^2+\frac{x}{\sqrt{2}}+1=0$	4	$\frac{-1 \pm i \sqrt{2 \sqrt{2}-1}}{2}$

• A. A= 2 ,B= 1 ,C= 4 ,D= 3
• B. A=1 ,B= 2,C= 3,D= 4
• C. A= 2 ,B=1 ,C=3 ,D=4
• D. A= 4 ,B=3 ,C=2,D= 1

Answer 1

Answer: (A)

A. Given, $\sqrt{2} x^2+x+\sqrt{2}=0$
On comparing the given equation with $a x^2+b x+c=0$, we get
$ a=\sqrt{2}, b=1, c=\sqrt{2} $
$\therefore D=b^2-4 a c=(1)^2-4 \times \sqrt{2} \times \sqrt{2}$
$ =1-4 \times 2=1-8=-7<0 $
$ \Rightarrow x =\frac{-1 \pm \sqrt{-7}}{2 \times \sqrt{2}} $
$ x =\frac{-1 \pm i \sqrt{7}}{2 \sqrt{2}} (\because \sqrt{-1}=i)$
B. Given, $\sqrt{3} x^2-\sqrt{2} x+3 \sqrt{3}=0$
On comparing the given equation with $a x^2+b x+c=0$, we get
$a=\sqrt{3}, b=-\sqrt{2}, c=3 \sqrt{3}$
Now, $ D=b^2-4 a c=(-\sqrt{2})^2-4 \times(\sqrt{3}) \times 3 \sqrt{3}$
$=2-4 \times 3 \times \sqrt{3} \times \sqrt{3}=2-12 \times 3$
$=2-36=-34<0$
$\Rightarrow x =\frac{-(-\sqrt{2}) \pm \sqrt{-34}}{2 \times \sqrt{3}}$
$ =\frac{\sqrt{2} \pm i \sqrt{34}}{2 \sqrt{3}} (\because \sqrt{-1}=i)$
C. Given, $x^2+x+\frac{1}{\sqrt{2}}=0$
On comparing the given equation with $a x^2+b x+c=0$, we get $a=1, b=1, c=\frac{1}{\sqrt{2}}$
Now, $D=b^2-4 a c=(1)^2-4 \times 1 \times \frac{1}{\sqrt{2}}$
$ =1-\frac{4}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}=1-\frac{4 \sqrt{2}}{2}=1-2 \sqrt{2}<0$
$\Rightarrow x =\frac{-1 \pm \sqrt{1-2 \sqrt{2}}}{2 \times 1}=\frac{-1 \pm i \sqrt{2 \sqrt{2}-1}}{2} (\because \sqrt{-1}=i)$
D. Given, $x^2+\frac{x}{\sqrt{2}}+1=0$
Taking the LCM to make the calculation easier,
$ \frac{x^2}{1}+\frac{x}{\sqrt{2}}+\frac{1}{1}=0, \frac{\sqrt{2} x^2+x+\sqrt{2}}{\sqrt{2}}=0 $
$ \Rightarrow \sqrt{2} x^2+x+\sqrt{2}=0$
On comparing the above equation with $a x^3+b x+c=0$, we get
$a =\sqrt{2}, b=1, c=\sqrt{2}$
$\text { Now, } D =b^2-4 a c=(1)^2-4 \times \sqrt{2} \times \sqrt{2}$
$ =1-4 \times 2=1-8=-7< 0 $
$\Rightarrow x =\frac{-1 \pm \sqrt{-7}}{2 \times \sqrt{2}}=\frac{-1 \pm i \sqrt{7}}{2 \sqrt{2}} (\because \sqrt{-1}=i)$