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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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3 Answers 3

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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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  • $\begingroup$ Thank you, that makes sense! $\endgroup$ Jun 18, 2013 at 0:51
  • $\begingroup$ would the solution be something like x=-3(+-)i sqrt3, all over 2? $\endgroup$ Jun 18, 2013 at 1:07
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    $\begingroup$ 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 $\endgroup$
    – amWhy
    Jun 18, 2013 at 1:12
  • $\begingroup$ Hmm. Ok, I'll have to go review this again. $\endgroup$ Jun 18, 2013 at 1:17
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    $\begingroup$ 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! $\endgroup$
    – amWhy
    Jun 18, 2013 at 1:25
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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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