Question

Quadratic systems

x^2+y^2=10
2y+y=4

the first part I got this x^2+(4-2x)^2=10
and I don't know what steps comes next.
could you show work on how to do it I'm so confuse on this problem.

Answer 1

Hmmm. By algebra 2 you should surely know how to solve quadratic equations...

x^2+(4-2x)^2=10
x^2+16-16x+4x^2-10 = 0
5x^2 - 16x + 6 = 0
...

Answer 2

I didn't learn this in class so could you tell me which step goes next. I want to learn how to do it and be able to do my homework by myself.

Answer 3

OK. But this is Algebra I, which you have presumably already passed. But, moving right along,

5x^2 - 16x + 6 = 0

The quadratic formula says that the roots of

ax^2+bx+c are

x = (-b±√(b^2-4ac)/2a

So, for this problem, that means

x = (16±√(16^2-4*5*6))/10
= (16±√136)/10
= (16±2√34)/10
= (8±√34)/5

Or, you can always solve by completing the square. Recall that

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

So, we have

5x^2 - 16x + 6 = 0
5x^2 - 16x = -6
5(x^2 - 16/5) = -6
5(x^2 - 16/5 + (8/5)^2) = -6 + 5(8/5)^2
5(x - 8/5)^2 = 34/5
x - 8/5 = ±√34/5
x = (8±√34)/5