Question

–5 is one of the zeroes of 2x^2 + px – 15, zeroes of p(x^2+ x) + k are equal to each other. Find the value of k. (DO NOT Beg For Brainliest , Or Else You Won't Get It...) ! 100 POINT SPECIAL QUESTION !

Answer 1

Answer:

k = -5/4

Step-by-step explanation:

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

2(25) +p(-5) -15 = 0

50 -5p -15 =0

-5p + 35 = 0

-5p = -35

p = -35/-5 = 7

-5x^2 -5x + k

as zeroes are equal

b^2 - 4ac = 0

(-5)^2 -4(-5)(k) = 0

25 + 20k = 0

20k = -25

k = -25/20 = -5/4

Answer 2

$\huge\purple{Answer}$

Given that (−5) is the root of 2x2 + px – 15 = 0

Put x = (−5) in 2x2 + px – 15 = 0

⇒ 2(−5)2 + p(−5) − 15 = 0

⇒ 50 −5p − 15 = 0

⇒ 35 − 5p = 0

⇒ 5p = 35

∴ p = 7

Hence the quadratic equation p(x2 + x) + k = 0 becomes, 7(x2 + x) + k = 0

⇒ 7 x2 + 7x + k = 0

Here a = 7, b = 7 and c = k

Given that this quadratic equation has equal roots

∴ b2 – 4ac = 0

⇒ 72 – 4(7)(k) = 0

⇒ 49 – 28k = 0

⇒ 49 = 28k

∴ k = (49/28) = 7/4