Question

Find the value of \[k\] for which the following equation has equal roots \[{x^2} + 4kx + {k^2} - k + 2 = 0\].

Answer 1

Hint:
Here, we will first compare the given equation with the general form of a quadratic equation to get the coefficient of \[{x^2}\], coefficient of \[x\], and the constant term. We will then use the condition of equal roots to find the Quadratic equation with the variable \[k\]. Then by using the quadratic roots formula, we will find the roots of the obtained quadratic equation and hence the value of \[k\].

Formula Used:
Quadratic roots is given by the formula \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]

Complete step by step solution:
We are given a quadratic equation \[{x^2} + 4kx + \left( {{k^2} - k + 2} \right) = 0\] which has equal roots.
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
\[\begin{array}{l}a = 1\\b = 4k\\c = {k^2} - k + 2\end{array}\]
We know that the Quadratic Equation has equal roots only if \[{b^2} - 4ac = 0\].
Now, by substituting the coefficient of \[{x^2}\], coefficient of \[x\] and the constant term in the above condition, we get
\[ \Rightarrow {\left( {4k} \right)^2} - 4\left( 1 \right)\left( {{k^2} - k + 2} \right) = 0\]
Simplifying the equation, we get
\[ \Rightarrow 16{k^2} - 4{k^2} + 4k - 8 = 0\]
Subtracting the like terms, we get
\[ \Rightarrow 12{k^2} + 4k - 8 = 0\]
Now, dividing the equation by 4, we get
\[ \Rightarrow 3{k^2} + k - 2 = 0\]
Now, we will find the roots of the above quadratic equation by using the quadratic roots formula.
By substituting the values \[a = 1,b = 4k\] and \[c = {k^2} - k + 2\] in the formula \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\], we get
\[k = \dfrac{{ - 1 \pm \sqrt {{1^2} - 4\left( 3 \right)\left( { - 2} \right)} }}{{2\left( 3 \right)}}\]
Now, by simplifying the equation, we get
\[ \Rightarrow k = \dfrac{{ - 1 \pm \sqrt {1 + 24} }}{6}\]
Adding the terms, we get
\[ \Rightarrow k = \dfrac{{ - 1 \pm \sqrt {25} }}{6}\]
By further simplifying the terms, we get
\[ \Rightarrow k = \dfrac{{ - 1 \pm 5}}{6}\]
Rewriting the above equation, we get
\[ \Rightarrow k = \dfrac{{ - 1 + 5}}{6}\] and \[k = \dfrac{{ - 1 - 5}}{6}\]
Adding and subtracting the terms, we get
\[ \Rightarrow k = \dfrac{{ + 4}}{6}\] and \[k = \dfrac{{ - 6}}{6}\]
Simplifying the expression, we get
\[ \Rightarrow k = \dfrac{2}{3}\] and \[k = - 1\]

Therefore, the value of \[k\] is \[\dfrac{2}{3}\] and \[ - 1\] for which the Quadratic Equation has equal roots.

Note:
We know that the Quadratic equation is an equation whose highest degree is 2. The Quadratic Equation can be easily solved by using the method of the 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.