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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.
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