Question

Find the value of k for which the quadratic equation $\left( {k - 2} \right){x^2} + 2\left( {2k - 3} \right)x + \left( {5k - 6} \right) = 0$ has equal roots.

Answer 1

Hint: In order to solve the following problem, we will use the property of determinant of the quadratic equation to find the relation between the coefficients of the quadratic equation when the roots are equal. We will find a quadratic equation in k using this property. After solving this quadratic equation we will find the value of k.

Complete step-by-step answer:
Given Data: $\left( {k - 2} \right){x^2} + 2\left( {2k - 3} \right)x + \left( {5k - 6} \right) = 0$ has equal roots.

We know for a polynomial equation of the form $a{x^2} + bx + c = 0$, the determinant D is given by $D = {b^2} - 4ac$

Here we get, $a = \left( {k - 2} \right),b = 2\left( {2k - 3} \right)\& c = \left( {5k - 6} \right)$ by comparing it with the standard polynomial $a{x^2} + bx + c = 0$

Determinant $D = {b^2} - 4ac$
$
   \Rightarrow D = {\left( {4k - 6} \right)^2} - 4\left( {k - 2} \right)\left( {5k - 6} \right) \\
   \Rightarrow D = 16{k^2} + 36 - 48k - 20{k^2} + 64k - 48 \\
   \Rightarrow D = - 4{k^2} + 16k - 12 \\
 $
We know the determinant D = 0, when quadratic equations have equal roots.
Therefore, we have:
$
  \because D = 0 \\
   \Rightarrow D = - 4{k^2} + 16k - 12 = 0 \\
   \Rightarrow - 4{k^2} + 16k - 12 = 0 \\
   \Rightarrow 4{k^2} - 16k + 12 = 0 \\
 $
Let us now solve the quadratic equation to find the value of k.
$
   \Rightarrow 4{k^2} - 16k + 12 = 0 \\
   \Rightarrow {k^2} - 4k + 3 = 0 \\
   \Rightarrow {k^2} - 3k - k + 3 = 0 \\
   \Rightarrow k\left( {k - 3} \right) - 1\left( {k - 3} \right) = 0 \\
   \Rightarrow \left( {k - 1} \right)\left( {k - 3} \right) = 0 \\
   \Rightarrow k = 1,k = 3 \\
 $
Hence, the value of k when the quadratic equation $\left( {k - 2} \right){x^2} + \left( {2k - 3} \right)x + \left( {5k - 6} \right) = 0$ has equal roots are 1 and 3.

Note: This problem can also be solved by first finding both of the roots of the quadratic equation given in terms of k and then equating them as given in the problem to find the value of k. But the relation of determinant for different conditions is a very important and easier way of solution. So, the key in solving such types of problems is to know that the quadratic equation has two equal real roots if the determinant equals 0. The relation of determinant for different conditions is as follows.

Nature of roots of a quadratic equation:
 If $D = {b^2} - 4ac > 0$ , the quadratic equation has two distinct real roots.
 If $D = {b^2} - 4ac = 0$ , the quadratic equation has two equal real roots.
 If $D = {b^2} - 4ac < 0$ , the quadratic equation has no real roots.