Question

Plz answer this fast Q.2)

Answer 1

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

Answer 2

HELLO DEAR,

given:-(x= -5)

2x²+px-15=0

now put the value of x in the Equation

we get,

2(-5)²+p(-5)-15=0

50 -5p -15=0

35=5p

p=35/5

p=7


now ,

given that the Equation {p(x²+x) +k =0 }
has a real and equal roots

{p(x²+x) +k =0 }

=>px²+px+k=0------------------(1)


where
a=p, b= p and ,c=k
we know that:-
$discriminent = \sqrt{ {b}^{2} - 4ac} = 0 \\ now \: \: \: put \: the \: values \: in \: the \: given \: equtions \: we \: get \\ = > \sqrt{ {p}^{2} - 4 \times p \times k} = 0 \\ = > \sqrt{ {7}^{2} - 4 \times 7 \times k } = 0 .........using.(1)\\ = > \sqrt{49 - 28k} = 0 \\ = > 49 = 28k \\ = > k = \frac{49}{28} \\ = > k = \frac{7}{4}$


I HOPE ITS HELP YOU DEAR,
THANKS