Question

The given quadratic equations have real roots and roots are Real and equal, $\sqrt{\dfrac{3}{2}:}$
$2{{x}^{2}}-2\sqrt{6x}+3=0$
(a) True
(b) False

Answer 1

Hint: Any quadratic $a{{x}^{2}}+bx+c=0$ will have real and distinct roots, if discriminant is greater than 0, roots will be real and equal if discriminant is 0 and will not have any real roots if discriminant is less than 0. Value of discriminant is given ass
$D={{b}^{2}}-4ac$
Roots of any quadratic $a{{x}^{2}}+bx+c=0$ are given as
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$

Complete step-by-step solution -
Given quadratic equation in the problem is
$2{{x}^{2}}-2\sqrt{6x}+3=0..........\left( i \right)$
As, we need to verify whether roots are real and equal or root and if equal then it is equal to $\sqrt{\dfrac{3}{2}}$ or not.
As we know the discriminant of any quadratic equation decides the property of roots i.e. whether the roots will be real or not will be decided by the discriminant of the quadratic.
We know the discriminant of any quadratic $a{{x}^{2}}+bx+c=0$ is given as
$D={{b}^{2}}-4ac..........\left( ii \right)$
And roots will be real and distinct if D > 0, roots will be real and equal, if D = 0 and roots will not be real (imaginary) if D < 0
So, let us calculate the discriminant of the quadratic given in equation(i) using the relation of the equation(ii). So, we get values of a, b, c by comparing the quadratic of equation(i) with the standard equation of quadratic i.e. $a{{x}^{2}}+bx+c=0.$ So, we get
$a=2,b=-2\sqrt{6},c=3$
Hence, discriminant can be given ass
$\begin{align}
  & D={{\left( -2\sqrt{6} \right)}^{2}}-4\times 2\times 3 \\
 & D=24-24 \\
 & D=0.........\left( iii \right) \\
\end{align}$
Now, as D = 0, then roots of the given equation will be real and equal.
Now, as we know the roots of any quadratic equation $a{{x}^{2}}+bx+c=0$ can be given as
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}........\left( iv \right)$
Where ${{b}^{2}}-4ac$ i.e. D can be replaced by 0 from the equation (iii). Hence, we get roots of the quadratic of equation(i) as
$x=\dfrac{-b}{2a}$
Now, we have
$b=-2\sqrt{6},a=2$
Hence, root of the quadratic is given as
$\begin{align}
  & x=\dfrac{-\left( -2\sqrt{6} \right)}{2\times 2} \\
 & x=\dfrac{2\sqrt{6}}{4}=\dfrac{\sqrt{6}}{2} \\
\end{align}$
Multiply the numerator and denominator by $\sqrt{2}$ , we get
$\begin{align}
  & x=\dfrac{\sqrt{6}}{2}\times \dfrac{\sqrt{2}}{\sqrt{2}} \\
 & x=\dfrac{\sqrt{12}}{2\sqrt{2}}=\dfrac{\sqrt{2\times 2\times 3}}{2\sqrt{2}} \\
 & x=\dfrac{2\sqrt{3}}{2\sqrt{2}}=\dfrac{\sqrt{3}}{\sqrt{2}}=\sqrt{\dfrac{3}{2}} \\
 & x=\sqrt{\dfrac{3}{2}} \\
\end{align}$
Hence, the root of the quadratic is $\sqrt{\dfrac{3}{2}}$ . Hence the quadratic in the problem has equal roots and equal to $\sqrt{\dfrac{3}{2}}$ .So, the given statement is true.
Hence, option(a) is correct.

Note: Another approach for the given question would be that we can write the given quadratic as
${{\left( \sqrt{2}x \right)}^{2}}-2\times \sqrt{2}x\times \sqrt{3}+{{\left( \sqrt{3} \right)}^{2}}=0$
Use identity ${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$ and get the above expression as
$\begin{align}
  & {{\left( \sqrt{2}x-\sqrt{3} \right)}^{2}}=0 \\
 & \Rightarrow \sqrt{2}x-\sqrt{3}=0 \\
 & x=\dfrac{\sqrt{3}}{\sqrt{2}}=\sqrt{\dfrac{3}{2}} \\
\end{align}$
So, it can be another approach to solve the problem.
Take care while substituting the values of a, b and c from the given quadratic in the problem to the formula of discriminant and roots as well. So, be careful with this step in the solution as well.