Question

If -4 is a root of the equation x² + px -4 =0and the Q.E x² + px + k =0 has equal
roots, find the value of k.
​

Answer 1

Given:

-4 is a root of the quadratic equation x²+px-4=0

ATQ,

x= -4

Implies,

x²+px-4=0

→(-4)²+p(-4)-4=0

→16-4=4p

→4p=12

→p=3

Also,

The quadratic equation x²+px+k=0 has real and equal roots. i.e.,D=0

Here,

x²+px+k=0»x²+3x+k=0

On comparing with ax²+bx+c=,

a=1,b=3 and c=k

Now,

D=0

→b²-4ac=0

→3²-4k=0

→k=9/4

Answer 2

Step-by-step explanation:

Hi,

P ( x ) = x² + PX - 4

P(-4) = (-4)² + P × -4 - 4 = 0

=> 16 - 4P - 4 = 0

=> -4P + 12 = 0

=> -4P = -12

=> P = 12/4 = 3.

Now,

X² + PX + k = 0

x² + 3x + k = 0

Here,

a = 1 , b = 3 and c = k

Discriminant = b² - 4ac

b² - 4ac = 0

3²- 4*1*k = 0

9 - 4k = 0

-4k = -9

k = 9/4

Hope it will help you :)