Question

Which of the following equations has two equal real roots?

• A. $2x^2-3\sqrt{2}x+\frac{9}{4}=0$
• B. $x^2+x+6=0$
• C. $x^2+3x+2\sqrt{2}=0$
• D. $5x^2+3x+1=0$

Answer 1

The correct option is A $2x^2-3\sqrt{2}x+\frac{9}{4}=0$

We know that for a quadratic equation, $a{x}^2+bx+c=0$
Discriminant, $D=b^2-4ac$
If $D=0$, roots are real and equal
If $D>0$, roots are real and distinct
If $D<0$, roots are imaginary and distinct

$\left(A\right)$ $2x^2-3\sqrt{2}x+\frac{9}{4}=0$

$\Rightarrow 8x^2-12\sqrt{2}x+9=0$

$D=b^2-4ac$

$=(12\sqrt{2}{)}^2-4\times 8\times 9$

$=288-288=0$

$\therefore D=0$

Hence, the roots of the equation will be real and equal.

$\left(B\right)$ $x^2+x+6=0$

$D=b^2-4ac$

$=(1{)}^2-4\times 1\times 6$

$=1-24$

$=-23<0$

$\therefore D<0$

Hence, the roots of the equation will be imaginary.

$\left(C\right)$ $x^2+3x+2\sqrt{2}=0$

$D=b^2-4ac$

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

$=9-8\sqrt{2}$

$=-2.313<0$

$\therefore D<0$

Hence, the roots of the equation will be imaginary.

$\left(D\right)$ $5x^2+3x+1=0$

$D=b^2-4ac$

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

$=9-20$

$=-11<0$

$\therefore D<0$

Hence, the roots of the equation will be imaginary.

Hence, only equation given in option A will have real and equal roots.