Question

For what value of $k$ the quadratic equation has equals roots, the equation is \[\left( {k - 2} \right){x^2} + 2\left( {2k - 3} \right)x + (5k - 6) = 0\]\[\left( {k - 2} \right){x^2} + 2\left( {2k - 3} \right)x + (5k - 6) = 0\] ?

Answer 1

Hint: The quadratic equation has been given to have equal roots. As a result, the determinants will be zero. We know that where a, b, and c are the coefficients of \[{x^2}\] , x, and constant terms of the provided quadratic equation in the question, respectively. To get the required value of k, solve the quadratic equation of k.

Complete step by step answer:
We know that if a polynomial equation's highest degree (highest power of the variable) is 2, the polynomial equation is quadratic. We also know that the universal quadratic equation looks like this:
\[ \Rightarrow a{x^2} + bx + c = 0\].........(1)
We know that the discriminant is defined as $D = b^2- 4ac$, where $a, b$ and $c$ are the coefficients of \[{x^2}\] , $x$ and constant terms respectively. Since, we have given that the equation has equal roots. So, the discriminant will be equal to zero.We will equate discriminant equals to zero. When We will compare equation \[\left( {k - 2} \right){x^2} + 2\left( {2k - 3} \right)x + (5k - 6) = 0\] with equation 1.

We get \[a = k - 2\], \[b = 4k - 6\], \[c = 5k - 6\]. We will put D equals to zero and the other values. We will put the value of a, b and c.
\[ \Rightarrow 0 = \;{\left( {{\mathbf{4k}} - {\mathbf{6}}} \right)^2} - {\mathbf{4}} \times \left( {{\mathbf{k}} - {\mathbf{2}}} \right)\left( {{\mathbf{5k}} - {\mathbf{6}}} \right)\]
We will expand the square term and multiply the second term
\[ \Rightarrow 0 = 16{k^2} + 36 - 48k - 20{k^2} + 64k - 48\]
\[ \Rightarrow 0 = 4{k^2} - 16k + 12\]
We will divide whole equation by 4
\[ \Rightarrow 0 = {k^2} - 4k + 3\]
We will use mid-term split formula
\[ \Rightarrow 0 = {k^2} - 3k - k + 3\]
\[ \Rightarrow 0 = \left( {k - 3} \right)\left( {k - 1} \right){\text{ }}\]
\[ \therefore k = 3\] and \[k = - 1\]

Hence, for k=3 and \[k = - 1\] the equation \[\left( {k - 2} \right){x^2} + 2\left( {2k - 3} \right)x + (5k - 6) = 0\] has equal roots.

Note: When we encounter this type of problem, we must first form the generic quadratic equation, which we will compare to the supplied equation to get the values of a, b, and c. Then, using the condition given in question, we find the equation in terms of k.