Question

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

Answer 1

Hint:
Here, we will use the condition of equal roots to find the Quadratic equation such that the variable will be \[k\]. Then by using the quadratic root formula, we will find the roots of the formed quadratic equation and hence the required value of \[k\]. A quadratic equation is an equation with the highest degree of variable as 2.

Formula Used:
We will use the following formula:
1) The square of the difference of two numbers is given by the Algebraic Identity \[{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab\].
2) Quadratic roots is given by the formula \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]

Complete step by step solution:
The given quadratic equation is \[\left( {k - 2} \right){x^2} + \left( {2k - 3} \right)x + \left( {5k - 6} \right) = 0\]
We know that the general Quadratic equation is of the form \[a{x^2} + bx + c = 0\]
By comparing the given Quadratic equation with the general Quadratic equation, we have
\[\begin{array}{l}a = k - 2\\b = 2k - 3\\c = 5k - 6\end{array}\]
It is given that the given Quadratic Equation has equal roots.
We know that the Quadratic Equation has equal roots only if \[{b^2} - 4ac = 0\].
Now, by using the condition of equal roots and by substituting the coefficient of \[{x^2}\], coefficient of \[x\]and the constant term in \[{b^2} - 4ac = 0\], we get
\[{\left( {2k - 3} \right)^2} - 4\left( {k - 2} \right)\left( {5k - 6} \right) = 0\]
Now, by using the Algebraic Identity \[{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab\], we get
\[ \Rightarrow \left( {{{\left( {2k} \right)}^2} + {{\left( 3 \right)}^2} - 2\left( 3 \right)\left( {2k} \right)} \right) - 4\left( {k - 2} \right)\left( {5k - 6} \right) = 0\]
By simplifying the equation, we get
\[ \Rightarrow \left( {4{k^2} + 9 - 12k} \right) - 4\left( {k - 2} \right)\left( {5k - 6} \right) = 0\]

Now, by multiplying the binomials by using the FOIL method, we get
\[ \Rightarrow \left( {4{k^2} + 9 - 12k} \right) - 4\left[ {k\left( {5k - 6} \right) - 2\left( {5k - 6} \right)} \right] = 0\]
\[ \Rightarrow \left( {4{k^2} + 9 - 12k} \right) - 4\left[ {5{k^2} - 6k - 10k + 12} \right] = 0\]
Again multiplying the terms, we get
\[ \Rightarrow \left( {4{k^2} + 9 - 12k} \right) - \left[ {20{k^2} - 24k - 40k + 48} \right] = 0\]
By simplifying the equation, we get
\[ \Rightarrow 4{k^2} + 9 - 12k - 20{k^2} + 24k + 40k - 48 = 0\]
Adding and subtracting the like terms, we get
\[ \Rightarrow - 16{k^2} + 52k - 39 = 0\]
Dividing by \[ - 4\] on both the sides, we get
\[ \Rightarrow 16{k^2} - 52k + 39 = 0\]
Now, by substituting the values in the quadratic roots formula \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\], we get
 \[k = \dfrac{{ - \left( { - 52} \right) \pm \sqrt {{{\left( { - 52} \right)}^2} - 4\left( {16} \right)\left( {39} \right)} }}{{2\left( {16} \right)}}\]
Simplifying the equation, we get
\[ \Rightarrow k = \dfrac{{52 \pm \sqrt {2704 - 2496} }}{{32}}\]
Subtracting the terms, we get
\[ \Rightarrow k = \dfrac{{52 \pm \sqrt {208} }}{{32}}\]
By simplifying the equation, we get
\[ \Rightarrow k = \dfrac{{52 \pm \sqrt {52 \times 4} }}{{32}}\]
\[ \Rightarrow k = \dfrac{{52 \pm 2\sqrt {52} }}{{32}}\]
By taking out the common factors, we get
\[ \Rightarrow k = \dfrac{{2\left( {26 \pm \sqrt {52} } \right)}}{{32}}\]
Dividing the terms, we get
\[ \Rightarrow k = \dfrac{{\left( {26 \pm \sqrt {52} } \right)}}{{16}}\]
Rewriting the equation, we get
\[ \Rightarrow k = \dfrac{{\left( {26 \pm \sqrt {13 \times 4} } \right)}}{{16}}\]
\[ \Rightarrow k = \dfrac{{\left( {26 \pm 2\sqrt {13} } \right)}}{{16}}\]
By taking out the common factors, we get
\[ \Rightarrow k = \dfrac{{2\left( {13 \pm \sqrt {13} } \right)}}{{16}}\]
Dividing the numerator and denominator by 2, we get
\[ \Rightarrow k = \dfrac{{\left( {13 \pm \sqrt {13} } \right)}}{8}\]

Therefore, the value of \[k\] is \[\dfrac{{13 + \sqrt {13} }}{8}\] and \[\dfrac{{13 - \sqrt {13} }}{8}\].

Note:
We know that some quadratic equations cannot be solved by using the factorization method and square root method. But whatever be the quadratic equation, it is quite easy to solve by using the method of quadratic formula. We should be careful that the quadratic equation should be arranged in the right form. We should also notice that we have both the positive and negative signs in the formula, so the solutions for the equations would be according to the signs. The values of the variable satisfying the given equation are called the roots of a quadratic equation.