Question

Question 9
Which of the following equation has two distinct real roots?
(a) $2x^2-3\sqrt{2}x+\frac{9}{4}=0$
(b) $x^2+x-5=0$
(c) $x^2+3x+2\sqrt{2}=0$​​​​​​​
(d) $5x^2-3x+1=0$​​​​​​​

Answer 1

a) Given equation is, $2x^2-3\sqrt{2}x+\frac{9}{4}=0$
On comparing with $a{x}^2+bx+c=0$
a = 2, b =$-3\sqrt{2}$ and c = $\frac{9}{4}$
$Now,D=b^2-4ac={\left(3\sqrt{2}\right)}^2-4\left(2\right)\left(\frac{9}{4}\right)=18-18=0$
Thus, the equation has real and equal roots.

b) The given equation is $x^2$ + x – 5 = 0
On comparing with $a{x}^2$ + bx + c = 0, we get
a = 1, b and c = - 5
The discriminant of $x^2$ + x – 5 = 0 is
$D=b^2-4ac=(1{)}^2-4(1\left)\right(-5)$
= 1 + 20 = 21
$\Rightarrow$ $b^2-4ac$ > 0
So, $x^2+x-5=0$ has two distinct real roots.

c) Given equation is $x^2+3x+2\sqrt{2}$ = 0
On comparing with $a{x}^2$ + bx + c = 0
a = 1, b = 3 and c = $2\sqrt{2}$
$now,D=b^2-4ac=(3{)}^2-4(1\left)\right(2\sqrt{2})=9-8\sqrt{2}<0$
$\therefore$ Roots of the equation are not real.

d) Given equation is, $5x^2$ – 3x + 1 = 0
On comparing with $a{x}^2$ + bx + c = 0
a = 5, b = - 3, c = 1
$Now,D=b^2-4ac=(-3{)}^2-4(5\left)\right(1)=9-20<0$
Hence, roots of the equation are not real.