Question

STATEMENT -1 : Roots of the quadratic equation $3x^2-2\sqrt{6}x+2=0$ are same.
STATEMENT -2 : A quadratic equation $a{x}^2+bx=c=0$ has two distinct real roots, if $b^2-4ac>0$

• A. Statement - 1 is True, Statement - 2 is True, Statement - 2 is a correct explanation for Statement - 1
• B. Statement - 1 is True, Statement - 2 is True : Statement 2 is NOT a correct explanation for Statement - 1
• C. Statement - 1 is True, Statement - 2 is False
• D. Statement - 1 is False, Statement - 2 is True

Answer 1

The correct option is B Statement - 1 is True, Statement - 2 is True : Statement 2 is NOT a correct explanation for Statement - 1

Given: quadratic equation $3x^2-2\sqrt{6}x+2=0$

To find the nature of the roots of the given equation.

Sol: The given equation is of the form $a{x}^2+bx+c=0$

Therefore, $a=3,b=-2\sqrt{6},c=2$

And for any quadratic equation,

(i) if Discriminant $b^2-4ac>0$, then the roots of the quadratic equation are real and unequal.

(ii) if Discriminant $b^2-4ac=0$, then the roots of the quadratic equation are real and equal.

(iii) if Discriminant $b^2-4ac<0$, then the roots of the quadratic equation are imaginary and unequal.

In the given equation the discriminant becomes

$b^2-4ac=(-2\sqrt{6}{)}^2-4(3\left)\right(2)=24-24=0$

Hence the given equation roots are same.

The roots can be found by using the formula,

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}⟹x=\frac{-(-2\sqrt{6})\pm \sqrt{(-2\sqrt{6}{)}^2-4(3\left)\right(2)}}{2\left(3\right)}⟹x=\frac{2\sqrt{6}}{6}⟹x=\frac{\sqrt{6}}{3},\frac{\sqrt{6}}{3}$