If $(2 p + 1) x^{2} - (p + 2) x + p - 1 = 0$ has repeated roots, then $7 p^{2} - 8 p =$

Open in App

Solution

Condition for repeated roots
$b^{2} = 4 a c$

$\Rightarrow (p + 2)^{2} = 4 (2 p + 1) (p - 1)$

$p^{2} + 4 p + 4 = 8 p^{2} - 8 p + 4 p - 4$

$p^{2} + 4 p + 4 = 8 p^{2} - 4 p - 4$

$7 p^{2} - 8 p - 8 = 0$

$7 p^{2} - 8 p = 8$

Suggest Corrections

Similar questions

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

p =

x^{2} - 2 p x + 8 p - 15 = 0

p =

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

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

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

p

Join BYJU'S Learning Program