# Choose the correct answer from the given four options in the following questions:Which of the following equations has two distinct real roots?(A) $2 x^{2}-3 \sqrt{2} x+\frac{9}{4}=0$(B) $x^{2}+x-5=0$(C) $x^{2}+3 x+2 \sqrt{2}=0$(D) $5 x^{2}-3 x+1=0$

To do:

We have to find the correct answer.

Solution:

$2 x^{2}-3 \sqrt{2} x+\frac{9}{4}=0$

Comparing with $a x^{2}+b x+c=0$, we get,

$a=2, b=-3 \sqrt{2}$ and $c=\frac{9}{4}$

$D=b^{2}-4 a c$

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

$=18-18$

$=0$

So, the equation has real and equal roots.

$x^{2}+x-5=0$

Comparing with $a x^{2}+b x+c=0$, we get,

$a=1, b=1$ and $c=-5$

$D=b^{2}-4 a c$

$=(1)^{2}-4(1)(-5)$

$=1+20$

$=21>0$

So, $x^{2}+x-5=0$ has two distinct real roots.
$x^{2}+3 x+2 \sqrt{2}=0$

Comparing with $a x^{2}+b x+c=0$

$a=1, b=3$ and $c=2 \sqrt{2}$

$D=b^{2}-4 a c$

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

$=9-8 \sqrt{2}<0$

So, the roots of the equation $x^{2}+3 x+2 \sqrt{2}=0$ are not real.

$5 x^{2}-3 x+1=0$

Comparing with $a x^{2}+b x+c=0$

$a=5, b=-3, c=1$

$D=b^{2}-4 a c$

$=(-3)^{2}-4(5)(1)$

$=9-20$

$=-11<0$

So, the roots of the equation $5 x^{2}-3 x+1=0$ are not real.

Updated on: 10-Oct-2022

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