Question

Find the value(s) of k so that the quadratic equation $3{{x}^{2}}-2kx+12=0$ has equal roots.

Answer 1

We will first find the roots of the given quadratic equation by using the formula $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$, and then equate them, since the roots are equal, to find all the possible values of k.
We know that the roots of a quadratic equation $a{{x}^{2}}+bx+c=0$ can be found by using the formula $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$.

Complete step-by-step solution:
The given equation is $3{{x}^{2}}-2kx+12=0$
Then the roots can be found by
$\begin{aligned} & x=\dfrac{-\left( -2k \right)\pm \sqrt{{{\left( -2k \right)}^{2}}-4\cdot 3\cdot 12}}{2\cdot 3} \\\ & =\dfrac{2k\pm \sqrt{4{{k}^{2}}-144}}{6} \\\ & =\dfrac{2k\pm \sqrt{4\left( {{k}^{2}}-36 \right)}}{6} \\\ & =\dfrac{2k\pm 2\sqrt{\left( {{k}^{2}}-36 \right)}}{6} \\\ & =\dfrac{k\pm \sqrt{\left( {{k}^{2}}-36 \right)}}{3} \end{aligned}$
The roots are $\dfrac{k+\sqrt{\left( {{k}^{2}}-36 \right)}}{3}$ and $\dfrac{k-\sqrt{\left( {{k}^{2}}-36 \right)}}{3}$.
It is known that the roots are equal.
Thus, equating these roots, we get

& \text{ }\dfrac{k+\sqrt{\left( {{k}^{2}}-36 \right)}}{3}=\dfrac{k-\sqrt{\left( {{k}^{2}}-36 \right)}}{3} \\\ & \Rightarrow k+\sqrt{\left( {{k}^{2}}-36 \right)}=k-\sqrt{\left( {{k}^{2}}-36 \right)} \\\ & \Rightarrow \sqrt{\left( {{k}^{2}}-36 \right)}=-\sqrt{\left( {{k}^{2}}-36 \right)} \\\ & \Rightarrow 2\sqrt{\left( {{k}^{2}}-36 \right)}=0 \\\ & \Rightarrow \sqrt{\left( {{k}^{2}}-36 \right)}=0 \\\ \end{aligned}$$ Squaring, we get $$\begin{aligned} & {{\left[ \sqrt{\left( {{k}^{2}}-36 \right)} \right]}^{2}}=0 \\\ & \Rightarrow {{k}^{2}}-36=0 \\\ & \Rightarrow {{k}^{2}}=36 \\\ & \Rightarrow k=\pm 6 \\\ \end{aligned}$$ **Hence, the value of k is $\pm 6$.** **Note:** Here, we can not solve this question by using the method of completing the square or by factorizing the middle terms into a product of two linear factors and then equating those linear factors to zero, since we do not know the numeric value of b, that is, it is a variable, in the given quadratic equation $3{{x}^{2}}-2kx+12=0$ of form $a{{x}^{2}}+bx+c=0$. Another way to solve this question is to directly put the value of discriminant as zero. We know that when the roots of a quadratic equation$a{{x}^{2}}+bx+c=0$ are equal, the value of $D={{b}^{2}}-4ac=0$. Hence, in this question, we can directly write that $\begin{aligned} & {{\left( -2k \right)}^{2}}-4\cdot 3\cdot 12=0 \\\ & \Rightarrow 4{{k}^{2}}-144=0 \\\ & \Rightarrow {{k}^{2}}-36=0 \\\ & \Rightarrow k=\pm 6 \\\ \end{aligned}$