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