Question

For what value of K, $K{x^2} - 5x + K = 0$ will have real and equal roots?

Answer 1

Hint: Here, we will proceed by comparing the given quadratic equation with any general quadratic equation $a{x^2} + bx + c = 0$ and then we will be using the formula that $d = \sqrt {{b^2} - 4ac} = 0$ for two real and equal roots.

Complete step-by-step answer:
Given, quadratic equation is $K{x^2} - 5x + K = 0{\text{ }} \to {\text{(1)}}$
For any general quadratic equation $a{x^2} + bx + c = 0{\text{ }} \to {\text{(2)}}$ to have real and equal roots, the value of the discriminant must be equal to zero where discriminant is given by $d = \sqrt {{b^2} - 4ac} $
By comparing the given quadratic equation given by equation (1) with the general quadratic equation given by equation (2), we get
a=K, b=-5 and c=K
For real and equal roots, $\sqrt {{b^2} - 4ac} = 0{\text{ }} \to {\text{(3)}}$
By substituting the values of a, b and c in equation (3), we have
$
   \Rightarrow \sqrt {{{\left( { - 5} \right)}^2} - 4\left( K \right)\left( K \right)} = 0 \\
   \Rightarrow \sqrt {25 - 4{K^2}} = 0 \\
   \Rightarrow 25 - 4{K^2} = 0 \\
   \Rightarrow 4{K^2} = 25 \\
   \Rightarrow {K^2} = \dfrac{{25}}{4} \\
   \Rightarrow K = \pm \sqrt {\dfrac{{25}}{4}} \\
   \Rightarrow K = \pm \dfrac{5}{2} \\
 $
Therefore, the values of K for which the given quadratic equation will have equal and real roots are $K = \dfrac{5}{2}$ and $K = - \dfrac{5}{2}$.

Note: In this particular problem, if we were asked for the value of the real and equal root for the given quadratic equation then in that case we will use the discriminant method for finding the roots of any general equation $a{x^2} + bx + c = 0$ which is $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$. Here, both the roots are real and equal so this formula is reduced to $x = \dfrac{{ - b}}{{2a}}$.