Question

Determine the value of k so that one root of x^2 + 8x + k = 0 is twice the other root.

Answer 1

Say the roots are \[x=r_1, r_2\]
We are given that one of the roots is twice the other.
So we have this:
\[x=r_1\]
\[x=r_2=2r_1\]
Ok so we have \[x=r_1 => x-r_1 \text{ is a factor}\]
\[x=2r_1 => x-2r_1 \text{ is another factor}\]
so when we factor \[x^2+8x+k\]
we are to get \[(x-2r_1)(x-r_1)\]
So what happens if we expand this (multiply these factors)
We get
\[(x-2r_1)(x-r_1)=x^2-r_1x-2r_1x+2r_1^2=x^2+x(-r_1-2r_1)+2r_1^2\]
But this is suppose to be equal to \[x^2+8x+k\]
So thats mean you have
\[-r_1-2r_1=8 \text{ and } 2r_1^2=k\]

Answer 2

\[-3r_1=8 => r_1=\frac{8}{-3}\]
\[2r_1^2=k => r_1=\pm \sqrt{\frac{k}{2}}\]
=>
\[\frac{8}{-3}=\pm \sqrt{\frac{k}{2}}\]

Answer 3

so we can say solve this for k

Answer 4

square both sides

Answer 5

\[\frac{64}{9}=\frac{k}{2}\]

Answer 6

not multiply both sides by 2

Answer 7

now*

Answer 8

\[\frac{128}{9}=k\]

Answer 9

i kept on messing up

Answer 10

(x-a) (x-2a) = x^2 - 3ax + 2a^2

x^2 + 8x + k

8 = -3a

a = 8/-3

k =128/9

Answer 11

yay! :)

Answer 12

@myininaya: Why don't you use vieta's formula?

Setting \( \alpha \) and \( 2 \alpha \) are the roots of the equation, then

$$ \alpha + 2 \alpha = -8 \Rightarrow \alpha = -\frac 83 $$

and,

$$ 2 \alpha^2 = k \Rightarrow k = \frac{ 128} {9} $$

and we are done!

Answer 13

i like my way because its longer lol

Answer 14

very nice fool

Answer 15

Thanks :D Simple is beautiful :)