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
...
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.
3 answers
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.
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
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
2 answers