Question

If -4 is the root of x² +px-4=0 and x² +px+q=0 has equal roots, find p and q​

Answer 1

In this equation, x²+px-4=0 one zero is 4.

Let the zeros be alpha and beta

We know that

$\alpha \beta = \frac{c}{a}$

Where a=1, b=p and c= -4

Therefore,

$( - 4) \beta = \frac{ - 4}{1} \\ = > \beta = 1$

Therefore the zeros of the quadratic equation x²+px-4=0 are 4 and 1

Answer 2

Answer:

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

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

Since -4 is the root for this equation, it can be equated to 0.

$0 = 16 - 4p - 4$

$0 = 12 - 4p$

$4p = 12$

$p = \frac{12}{4}$

$p = 3$

Applying the value of p in the second equation, we get:

${x}^{2} + px + q$

${x}^{2} + 3x + q$

Also, it is given in there question that the roots of this polynomial are equal, thus: discriminant must equal to zero.

Firstly, comparing the given equation to

$a {x}^{2} + bx + c$

We obtain the values of a, b, and c as follows:

a = 1

b = 3

c = q

Applying these values to the discriminant, we get:

$D = {b}^{2} - 4ac$

Since the roots are equal, the discriminant can be equated to zero.

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

$9 - 4q = 0$

$9 = 4q$

$q = \frac{9}{4}$

Hope this helps.

Thanks.