The value of $p > 0$ , for which both the equations $x^{2} + p x + 64 = 0$ & $x^{2} - 8 x + p = 0$ have equal roots is

8

No worries! We‘ve got your back. Try BYJU‘S free classes today!

16

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

32

No worries! We‘ve got your back. Try BYJU‘S free classes today!

64

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $16$
Given equation $x^{2} + p x + 64 = 0$
For equal roots $D = 0$
$\Rightarrow p^{2} - 4 (64) = 0$
$\Rightarrow p = \pm 16$
But it is given that $p > 0$
$\Rightarrow p = 16$
Other equation is $x^{2} - 8 x + p = 0$
For equal roots $D = 0$
$64 - 4 p = 0$
$p = 16$
Hence, $p = 16$ .

Suggest Corrections

Similar questions

x^{2} + p x + 64 = 0

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

x^{2} + 2 p x + 64 = 0 a n d X^{2} - 8 x + 2 p = 0

K

x^{2} + k x + 64 = 0

x^{2} - 8 x + k = 0

α, β

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

α^{3}, β^{3}

Join BYJU'S Learning Program