If $x^{2} - (5 m - 2) x + 4 m^{2} + 10 m + 25 = 0$ were to be a perfect square then find the value of $m$ ?

Open in App

Solution

$x^{2} - (5 m - 2) x + 4 m^{2} + 10 m + 25 = 0$ is a perfect square
$\Rightarrow D = 0$
$D = b^{2} - 4 a c$
${(5 m - 2)}^{2} - 4 (4 m^{2} + 10 m + 25) = 0$
$25 m^{2} + 4 - 20 m - 16 m^{2} - 40 m - 100 = 0$
$\Rightarrow 9 m^{2} - 60 m - 96 = 0$
$\Rightarrow 3 m^{2} - 20 m - 32 = 0$
$m = \frac{20 \pm \sqrt{400 + 384}}{6} = \frac{20 \pm 28}{5}$
$∴ m = 8, \frac{- 4}{3}$

Suggest Corrections

Similar questions

x^{2} - (5 m - 2) x + (4 m^{2} + 10 m + 25)

m

\sqrt{9 x^{4} + 6 x^{3} - 5 x^{2} - m x + 1}

m

(x + 1) (x + 2) (x + 3) (x + k) + 1

x^{2} + \frac{1}{25} x^{2} y^{2}

m

x^{2} - 2 (5 + 2 m) x + 3 (7 + 10 m) = 0

Join BYJU'S Learning Program