Question

If one root of the equation x^2+kx-8=0 is the square of the other, what is k?

A.) -6
B.) -2
C.) 2
D.) 6
E.) 8

Answer 1

x=k ± − k2+32−−−−−−√ 2

Answer 2

so can we take a, a^2 as the 2 roots ?

Answer 3

So the answer would be C

Answer 4

@erika_riera

Answer 5

I put the answer c the first time but I got it wrong.

Answer 6

Well try B

Answer 7

Hold on

Answer 8

why are you ppl guessing instead of solving? :P

Answer 9

Yeah thats why i said hold on ^^^:P

Answer 10

I'm doing test corrections and I need someone to explain to me how to solve it

Answer 11

\(\color{#0cbb34}{\text{Originally Posted by}}\) @hartnn
so can we take a, a^2 as the 2 roots ?
\(\color{#0cbb34}{\text{End of Quote}}\)

Answer 12

yes

Answer 13

product of roots for \(ax^2+bx+c = 0\)
is = c/a

so product of roots in your case = a*a^2 = -8/1

does this make sense?

Answer 14

yes I get that

Answer 15

a^3 = -8
find a

Answer 16

where did you get the ^3 from?

Answer 17

\(\Large a \times a^2 = a^{1+2} = a^3\)

Answer 18

a=2?

Answer 19

2^3 = 8
not -8

Answer 20

a=-2?

Answer 21

yes!

Answer 22

so a = -2 is one of the root
now you can just plug in
x= -2 in your original equation to get k

Answer 23

x^2+kx-8=0

(-2)^2 -2k - 8 =0
get k

Answer 24

So k=-2?

Answer 25

yes!

Answer 26

oh okay I get it now thank you!

Answer 27

welcome ^_^