Find the value of $k$ if the equation $k^{2} x^{2} - 2 (2 k - 1) x + 4 = 0$ has equal roots.

Open in App

Solution

Step 1: Determining the discriminant $(D)$ of the quadratic equation
Given that, equation $k^{2} x^{2} - 2 (2 k - 1) x + 4 = 0$ has equal roots.
The discriminant of the quadratic equation is determined by using the formula:
$D = B^{2} - 4 A C$
For equation $k^{2} x^{2} - 2 (2 k - 1) x + 4 = 0$ the value $A = k^{2}, B = - 2 (2 k - 1)$ and $C = 4$ .
Substituting the values of $A$ , $B$ and $C$ in discriminant formula.
$\begin{array}{rcl} D & = & B^{2} - 4 A C \\ = & {[- 2 (2 k - 1)]}^{2} - 4 (k^{2}) (4) \\ = & 4 (4 k^{2} - 4 k + 1) - 4 (4 k^{2}) \\ = & 4 (4 k^{2} - 4 k + 1 - 4 k^{2}) \\ = & 4 (- 4 k + 1) \end{array}$
Step 2: Determining the value of $k$
According to the nature of roots of a quadratic equation if the discriminant $D = 0$ , then the quadratic equation has real and equal roots.
By nature of roots of equation
$\begin{array}{rcl} D & = & 0 \\ \Rightarrow & 4 [- 4 k + 1] = 0 \\ \Rightarrow & - 4 k = - 1 \\ \Rightarrow & k = \frac{1}{4} \end{array}$
Hence, for $k = \frac{1}{4}$ the equation $k^{2} x^{2} - 2 (2 k - 1) x + 4 = 0$ has equal roots.

Suggest Corrections

Similar questions

x^{2} + 2 x + 4 p = 0

x^{2} + p x (1 + 3 k) + 7 (3 + 2 k) = 0

x^{2} + 4 x k x + k^{2} - k + 2 = 0.

x^{2} - 2 k x - 2 x + k^{2} = 0

k

x^{2} - 2 (k + 1) x + k^{2} = 0

k

x^{2} - 2 (k + 2) x + 12 + k^{2} = 0

Join BYJU'S Learning Program