Question

if -4 is the root of quadratic equation x²+kx-4=0and the quDratic equation x²+px+k then the equal roots to find the value of p and k​

Answer 1

Hey mate!

Here's your answer!

Since -4 is the root of quadratic equation x²+kx-4 = 0, it will satisfy the quadratic equation. So,

$( - 4) {}^{2} + k( - 4) - 4 = 0 \\ = > 16 - 4k - 4 = 0 \\ = > 12 - 4k = 0 \\ = > k = 3$

When k = 3 and x = -4, then x²+px+k = 0 is,
(-4)²+p(-4)+3 = 0
=> 16-4p+3 = 0
=> 19-4p = 0
=> p = 19/4
Hence the values of k and p are 3 and 19/4 respectively.

Hope it helps :)

Answer 2

Hi friend

If -4 is the root of x²+kx-4=0,we put -4 to corresponding values for x

x²+kx-4=0

=(-4)²+k(-4)-4=0

=16-4k-4=0. (1)

Put the value -4 in equation x²+px+k

=(-4)²+p(-4)+k=0

=16-4p+k=0. (2)

(1)-(2)

16-4k+4=0

- 16-4k-4p=0

=4p+4=0

=4(p+1)=0

=p+1=0

=p=-1

Now put the value of p in eq.(2) We get,

16-4(-1)-4k=0

=16+4-4k=0

=20-4k=0

4k=20

=k=5.

Therefore; p=-1 and k=5.

Please mark my answer as brainliest.Thank you.