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looking for help on this question. Solve the following systems of equations algebraically using the quadratic formula.

$$\begin{align} y& =-x^2+2x+9\\ y& =-5x^2+10x+12\end{align}$$

Any help would be appreciated!

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Is this homework? IF so use the homework tag. IT is best to tell us SE lurkers what you have tried to do to solve the problem. That way we can give an intelligent hint without giving the whole solution. –  ncmathsadist Jun 18 '13 at 0:52

3 Answers 3

up vote 7 down vote accepted

Hint: put both equations equal to one another. You'll have a quadratic equation.

$$-x^2+2x+9 = -5x^2+10x+12$$

Now simplify (combine "like terms": get all terms on one side (say left hand side) equal to $0$ (on the right-hand side).

Factor and/or use the quadratic formula to find any solution(s), if they exist, and they do exist: there are two solutions for $x$, each of which is a "zero" of the equation.

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Thank you, that makes sense! –  ComradeYakov Jun 18 '13 at 0:51
    
would the solution be something like x=-3(+-)i sqrt3, all over 2? –  ComradeYakov Jun 18 '13 at 1:07
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No, you should get $\dfrac{8 \pm \sqrt{64 - (-48)}}{8}$, which can be simplified. Check your simplification again, and make sure you are using the Quadratic Formula –  amWhy Jun 18 '13 at 1:12
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Yes! $x = \dfrac{2\pm \sqrt 7}{2}$ so $x_1 = 1 + \dfrac{\sqrt 7}2, \; x_2 = 1 - \dfrac{\sqrt 7}{2}$. Good work! –  amWhy Jun 18 '13 at 1:25
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@Jake Don't forget that you can click the green "check" mark on amWhy's answer to "accept" it. :) –  anorton Jun 18 '13 at 1:34

set $z=-x^2+2x$, we get:

$y=z+9\\y=5z+12$

So, $z=-\frac{3}{4}, y=\frac{33}{4}$

Then you can solve $-\frac{3}{4}=-x^2+2x$.

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Note: Because the system you are given has $y$ solved for in both cases, graphing could be an option.

However, this does not excuse one from knowing how to solve the system algebraically.

Solving Graphically: \begin{align} y& =-x^2+2x+9\\ y& =-5x^2+10x+12\end{align} https://www.desmos.com/calculator/lkub3kc0hx

Looking for intersecting points we find $x_1=-.32$ & $x_2=2.32$

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