Question

if (-4) is a root of the quadratic equation x^2 + px - 4 = 0 and the quadratic equation x^2 + px +k = 0 has equal roots/ find the value of k

Answer 1

Answer:

k = 9/4

Step-by-step explanation:

x^2 + px - 4 = 0

-4 is the root of this polynomial,

put x = -4

-4^2 + p(-4) - 4 = 0

16-4p-4 = 0

12 = 4p

p = 3

x^2 + px + k = 0

put p = 3

x^2 + 3x + k = 0

a = 1, b = 3, c = k

The above polynomial has equal roots,

D = 0

√b^2-4ac = 0

√9-4k = 0

9-4k = 0

9 = 4k

K = 9/4

Answer 2

Answer:

$\blue{k=9/4}$

Step-by-step explanation:

$\boxed{given}$

-4 is a root of the quadratic equation

$\underline{x^2+px-4}$

and the quadratic equation $\underline{x^2+px+k}$ has equal roots

$\boxed{Tofind}$

value of k

$\boxed{Answer}$

we have to find the value of p

that is

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

$( - 4) {}^{2} + p( - 4) - 4 = 0$

$16 - 4p - 4 = 0$

$12 - 4p = 0$

$- 4p = -12$

$p = 3$

now in the quesion the qudratic equation

$\underline{x^2+px+k=0}$

has equal root

so the discriminant will be 0

now we have

$b {}^{2} - 4ac = 0 \: for \: equal \: roots$

$b = value \: of \: p = 3$

$a = 1$

$c = k(we \: have \: to \: find)$

$(3) {}^{2} - 4(1)(k) = 0$

$9 - 4k = 0$

$- 4k = - 9$

$k = \frac{ - 9}{ - 4}$

$\blue{k=9/4}$