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!

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!

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

It shouldn't be that complicated.

anonymous · Answer

No it is not

anonymous · Answer

$$2x^2-8x=0,2x(x-4)=0$$
either x=0,
or x-4=0
x=4

anonymous · Answer

Oh so you basically "foiled it" I see.

anonymous · Answer

$$\frac{ y^2 }{7 }=y$$

anonymous · Answer

Factored

anonymous · Answer

multiply by 7 and solve like above question.