Question

2x^2 = 8x

y^2/7 = y

Solve for x on the first, and solve for y on the other. They're separate problems. I need to study for a test, and don't know how to do these, so if you could show work, that'd be great!

Answer 1

So. we have this:
\[2x^{2}=8x\]
Now substract 8x from both sides and we will have \[2x^{2}-8x=8x-8x\]:
\[2^{2}x-8x=0\]

Answer 2

So far I'm understanding it. so what's x? Or do we substitute 0 for x

Answer 3

This is has the form:
\[ax^{2}+bx+c=0\]
where a is the coefficient that multiplies x^2, b is the coefficient of x and c is the constant

Answer 4

And you can solve it with the quadratic formula:
\[x=\frac{ -b \pm \sqrt{b^{2}-4ac} }{ 2a }\]

Answer 5

It shouldn't be that complicated.

Answer 6

No it is not

Answer 7

\[2x^2-8x=0,2x(x-4)=0\]
either x=0,
or x-4=0
x=4

Answer 8

Oh so you basically "foiled it" I see.

Answer 9

\[\frac{ y^2 }{7 }=y\]

Answer 10

Factored

Answer 11

multiply by 7 and solve like above question.