Question

If (4-k )x² + 2(k +2 )x + (8k + 1 ) = 0 has equal roots, then k is​

Answer 1

       $\underline{\bold{Solution:}}$

       $\text{The given quadratic equation is }(4-k)x^{2} + 2(k+2)x + (8k+1) = 0 \text{ has}\\\text{equal roots}$

$\implies \text{The discriminant}(\delta) \text{ of the equation} = 0$

       $\text{We know that the discriminant}(\delta) \text{ of an equation of the form }ax^{2} +bx + c \\\text{is }b^{2} - 4ac$

       $\text{In our case:}$

       $a = (4-k)$

       $b = 2(k+2)$

       $c = (8k+1)$

$\implies \delta = [2(k+2)]^{2} - 4(4-k)(8k+1) = 0$

       $\text{Simplifying...}$

$\implies \delta = 4(k+2)^{2} - 4(4-k)(8k+1) = 0$

       $\text{We know that }(a+b)^{2} = a^{2} + b^{2} + 2ab$

$\implies \delta = 4(k^{2} + 4 + 4k) - 4(32k + 4 - 8k^{2} - k) = 0$

       $\text{Simplifying...}$

$\implies \delta = 4(k^{2} + 4 + 4k) - 4( - 8k^{2} +31k + 4) = 0$

       $\text{Taking 4 common and transposing it to R.H.S}$

$\implies \delta = k^{2} + 4 + 4k + 8k^{2} -31k - 4 = 0$

       $\text{Simplifying...}$

$\implies 9k^{2} -27k = 0$

       $\text{Taking }k \text{ common}$

$\implies 9k(k -3) = 0$

$\implies \boxed{k = 0,3}$