Going through my substitution section of my homework, and I successfully did most of them but had trouble with two of them:
Determine the intersection of the following systems:
a) y = 2x – 7 and y = - x^{2} -4 x – 3
Heres my attempt:
(2x-7)=-x^{2} -4x-3
0=x^{2 }-4x-3-2x+7
-1(0=-x^{2 }-6x+4)
0=x^{2}+6x-4
In every other question we've had we could just simplify this last step to end up with an answer like (x-x)(x-x), however with this one I can't see what multiplies to -4, and adds to 6
Then there's the second one
b) x^{2} + y^{2} = 30 and 2 x – 3 y = 10
2x-3y=10 ~ y= 2x + 10
3 3
x^{2 }+(2x + 10)^{2} = 30
.........3......3 [sorry about dots, still learning how to type in equations]
Is this correct so far? It doesn't feel like it is....
For quadratic equations you can find the zeros by the following methods (in order of decreasing difficulty): Rational Root Theorem, Factoring, Completing the Square, Quadratic Equation.
It's useful to know all 4.
For problem 1, completing the square you arrive at +/- sqrt(13)-3
Problem 2, you seem to be doing it right. Squaring that linear expression gives you a trinomial of the 2nd degree. Combine terms. Use quadratic equations if it's messy.
Please refer to our LaTeX forum. Basic use is quick and much easier for us to understand what equations you are trying to type. Look up \sqrt and \frac.
-Dan
topsquark I will check that out tomorrow, I have to go to bed right now, so I'm not going to be able to do it now.
For now I'll just clean it up by posting a picture, hopefully if someone answers I can check it in the morning before my test.
Can anyone solve this for me? This is a sample question for my quiz tomorrow that I'm still not ready for, could really, really use an answer.