Question

If (-5) is a root of the quadratic equation 2x2 + px - 15 = 0 and the quadratic equation p (x2 + x ) + k = 0 has equal roots, then find the values of p and k.

Answer 1

The first quadratic equation given is :-

$2 {x}^{2} + px - 15 = 0$

Also,

one root of the equation is (-5)

So,

According to remainder theorem :-

P(-5)=0

So,

$2{( - 5)}^{2}+ p ( - 5) - 15 = 0\\\\50 - 5p - 15 = 0\\\\5p = 35\\\\ p =\frac{35}{5}=7$

Thus,

Value of p is 7.

Now ,

The second quadratic equation is

$p({x}^{2}+ x) + k = 0\\\\7{x}^{2}+ 7x + k = 0$

The equation has equal roots.

This means that :-

D=0

So,

D = b²-4ac

So,

Value of D is equal to :-

$d = {7}^{2} - 4.7.k\\\\49 - 28k = 0\\\\28k = 49\\\\4k = 7\\\\k =\frac{7}{4}$

So,

Value of k is 7/4

Answer 2

$\bf\huge\boxed{\boxed{\bf\:Hello\:Mate}}}$

Given = (−5) is the root of 2x^2+ px – 15 = 0

Substitute value of  x = (−5)

⇒ 2(−5)^2 + p(−5) − 15 = 0

⇒ 50 − 5p − 15 = 0

⇒ 35 − 5p = 0

⇒ 5p = 35

⇒ p = 35 ÷ 5

→ Hence :-

p = 7

→ Hence :-

⇒ 7x^2 + 7x + k = 0

We get a = 7, b = 7 and c = k

Quadratic equation has equal roots

b^2 – 4ac = 0

⇒ 72 – 4(7)(k) = 0

⇒ 49 – 28k = 0

⇒ 49 = 28k

k = 49 ÷ 28

k = 7 ÷ 4

k = 1.75

$\bf\huge\boxed{\boxed{\bf\:P\:=\:7\:and\:k\:=\:7/4}}}$