Question

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

Answer 1

If -5 is the root of the polynomial:

P(X) = 2x²+px-15 = 15

2(-5)²+P*-5-15 = 15

50-5P = 30

P = 20/5

P = 4

p(X) = p(x^2+x)k

= pkx²+pkx + 0

For equal roots = b²-4ac = 0

(pk)² - 4*pk*0 = 0

(4k)² - 0 = 0

16k² = 0

k = √0

k = 0

Answer 2

Answer:

Step-by-step explanation:

P(x) =2(-5) ^2+p(-5) -15=0

2(25) -5p-15=0

50-5p-15=0

35-5p=0

P=35/5

P=7

P(x^2+x) +K=0

7(x^2+x) +k=0

7x^2+7x+k=0

Use b^2-4ac=0

(7) ^2-4(7) (k) =0

49-28k=0

K=49/28

=7/4

Ans