# Which of the following equations

Question:

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

Solution:

(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.