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) $5 x^{2}-3 x+1=0$
(b) The given equation is $x^{2}+x-5=0$
On comparing with $a x^{2}+b x+c=0$, we get
$a=1, b=1$ and $c=-5$
The discriminant of $x^{2}+x-5=0$ is
$D=b^{2}-4 a c=(1)^{2}-4(1)(-5)$
$=1+20=21$
$\Rightarrow \quad b^{2}-4 a c>0$
So, $x^{2}+x-5=0$ has two distinct real roots.
(a) Given equation is, $2 x^{2}-3 \sqrt{2} x+9 / 4=0$.
On comparing with $a x^{2}+b x+c=0$
$a=2, b=-3 \sqrt{2}$ and $c=9 / 4$
Now, $D=b^{2}-4 a c=(-3 \sqrt{2})^{2}-4(2)(9 / 4)=18-18=0$
Thus, the equation has real and equal roots.
(c) Given equation is $x^{2}+3 x+2 \sqrt{2}=0$
On comparing with $a x^{2}+b x+c=0$
$a=1, b=3$ and $c=2 \sqrt{2}$
Now, $D=b^{2}-4 a c=(3)^{2}-4(1)(2 \sqrt{2})=9-8 \sqrt{2}<0$
$\therefore$ Roots of the equation are not real.
(d) Given equation is, $5 x^{2}-3 x+1=0$
On comparing with $a x^{2}+b x+c=0$
$a=5, b=-3, c=1$
Now, $D=b^{2}-4 a c=(-3)^{2}-4(5)(1)=9-20<0$
Hence, roots of the equation are not real.
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