$$y=-x^2+2x+9$$ $$y=-5x^2+10x+12$$ Round answer to two decimal places.

So far I made both equations equal the other which lead to $-4x^2+8x+12$, took $4$ out, $4 (-x^2+2x+3)$. Then that bracketed terms were put into the quadratic formula to equal $(-1 \pm \sqrt{2})/(-1)$

That answer doesn't seem right, but I'm not sure where I went wrong.

Quadratic formula :

$$x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$$

  • $\begingroup$ Out of interest, what do you think the quadratic formula is? Because $b^2-4ac=4-4*(-1)*3=16$, so not sure where you got $\sqrt 2$ from $\endgroup$
    – mdave16
    Nov 10 '17 at 23:10
  • $\begingroup$ I edited your answer, hopefully, I didn't make any errors. I did put parentheses that were missing around the numerator in the quadratic formula - was this what you were missing? $\endgroup$ Nov 10 '17 at 23:10
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    $\begingroup$ You said that the equations lead to $-4x^2+8x+12$, but this is only an expression, not an equation. Perhaps you meant $-4x^2+8x+12=0$? I know that this doesn't look like much, but it makes a big difference. $\endgroup$ Nov 10 '17 at 23:11
  • $\begingroup$ @Michael Burr Yes, that's what I mean. Thank you. $\endgroup$
    – Grimestock
    Nov 11 '17 at 0:07
  • $\begingroup$ I got $(-2 \pm \sqrt{7})/(-2)$ as an answer. $\endgroup$
    – Grimestock
    Nov 11 '17 at 0:43

we have $$y=-x^2+2x+9$$ and $$y=-5x^2+10x+12$$ eliminating $y$ we have $$-5x^2+10x+12=-x^2+2x+9$$ and from here we get a quadratic equation $$0=4x^2-8x-3$$ using the quadratic fomula we obtain: $$x_1=\frac{2+\sqrt{7}}{2}$$ or $$x_2=\frac{2-\sqrt{7}}{2}$$

  • $\begingroup$ Thank you! how would I round these answers to two decimal places? Would it be -0.32 and 2.32?And I got -2 in my answer, how did you get positive 2? $\endgroup$
    – Grimestock
    Nov 11 '17 at 5:28
  • $\begingroup$ what you have : $(-2 \pm \sqrt{7})/(-2)$ is very essentially the same exact thing as $(2 \pm \sqrt{7})/2$. Just cancel the negative on the top and bottom, or if you like, multiply the numerator and the denominator by $-1$. I suspect that the quadratic you used was the exact same as the one in this answer multiplied by $-1$. Also, yes if they say round to two decimal places your answers meet those requirements. $\endgroup$ Nov 13 '17 at 18:45

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