Question

For what values of K the quadratic equation \[{{a}^{2}}-Ka+1=0\] has equal roots?

Answer 1

Hint: We will use the concept here that a quadratic equation has real and equal roots if and only if the discriminant of the equation is equal to zero. So, for this equation we may equate the discriminant to be equal to zero and get the value of K.

Complete step-by-step answer:
Since, the given quadratic equation is:
\[{{a}^{2}}-Ka+1=0\].
We know that the formula for finding the determinant of a quadratic equation $a{{x}^{2}}+bx+c=0$ is given as:
$D={{b}^{2}}-4ac$
Here, a is the coefficient of ${{x}^{2}}$, b is the coefficient of x and c is the constant term.
Therefore, according to this formula discriminant of the given equation is:
$D={{\left( -K \right)}^{2}}-4\times 1\times 1={{K}^{2}}-4$
Since, we have the condition that for the roots to be real and equal, the value of the discriminant must be equal to zero. So, on equating the above value of discriminant to zero, we get:
$\begin{align}
  & {{K}^{2}}-4=0 \\
 & {{K}^{2}}-{{2}^{2}}=0 \\
 & \left( K-2 \right)\left( K+2 \right)=0 \\
\end{align}$
So, here we will get two values of K which are K= 2 or K=-2.
Hence, the values of K for the given quadratic equation to have equal roots are 2 and -2.

Note: Here, it is important to note that a quadratic equation can have equal and real roots only when the discriminant of the quadratic equation is equal to zero. We should use the formula of discriminant of a quadratic equation carefully to avoid the mistakes and so that we get the correct value of K.