Question

x^2+px+q = (x-4) (x+5) find the value of p​

Answer 1

−4 is the root of x

2

+px−4=0

⇒(−4)

2

+p(−4)−4=0

⇒16−4p−4=0

⇒12=4p

so p=3

x

2

+px−q=0

⇒x

2

+3x+q=0

It has equal roots so D=0

b

2

−4ae=0

9−4q=0

so q= ⁹/4

Answer 2

Given:--

x²+px+q=(x-4)(x+5)

Start solving RHS(Right hand side)--

$\mathtt \red{ = (x - 4)(x + 5) }\\ \: \blue{ = {x}^{2} + 5x - 4x - 20}$

$\green{( {x}^{2} + x - 20) = ({x}^{2} + px + q)}$

Comparing both equations:

$1) \underbrace \mathtt{px = x }\\ \boxed {\orange{p = 1}}$

$2) \underbrace \mathtt{q = - 20 }\\ \boxed {\purple{q = - 20}}$

$\star\pink{{plz \: mark \: brainliest \: and \: thanks} }\star$