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 k

Answer 1

p(-5)=0=2(-5)^2+(-5)p-15
0=50-15-5p
5p=35
p=7
f(x)=7(x^2+x)+k=0
7x^2+7x+k=0
As roots are real and equal;
Therefore;
0=D
b^2-4ac=0
b^2=4ac
(7)^2=4×7×k
49=28k
7=4k
k=7/4

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