Question

if -4 is a root of the quadratic equation x^2 + Px - 4 =0 and x^2 + Px + k =0 has equal roots, find the value of k.

Answer 1

Answer:

K=-4

Step-by-step explanation:

as the root is -4 we can put the value of x = -4 and solve the equation

$f1(x) = {x}^{2} + p(x) - 4 = 0 \\ = f1( - 4) = {( - 4)}^{2} + p( - 4) - 4 = 0 \\ = 16 - 4p - 4 = 0 \\ = 12 - 4p = 0 \\ = 12 = 4p \\ = p = 3 \\ \\ now \: putting \: the \: value \: of \: x \: and \: p \: we \: get \: the \: value \: of \: \\ f2(x) = {x}^{2} + px + k = 0 \\ f2( - 4) = {( - 4)}^{2} + p( - 4) + k = 0 \\ = 16 - 4p + k = 0 \\ = 16 - (4 \times 3) + k = 0 \\ = 16 - 12 + k = 0 \\ = 4 + k = 0 \\ k = - 4$

Answer 2

Step-by-step explanation:

I).x^2+px-4=0

also, x=-4 (given)

so, (-4)^2+p×(-4) -4=0

=> 16-4 -4p=0

=>4p=12

==> p=3

Now,

II).x^2+px+k=0

also,

Discriminate =0

so,

{D= b^2-4ac}

So,

=> p^2-4×1×k=0

=> 9 -4k=0 (because p= 3 )

=> so, k= 9/4

answer.......

thank you.........