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“The points of intersection of the graph of $y=3+x-0.5x^2$ and the line $y=k$ are the solutions of the equation $10+2x-x^2=0$”

I was thinking that maybe i could find the solutions for the second equation and then find value of $k$. But thats too much work for one mark, and anyway it says to use the line $y=k$ to find the solutions. So i am supposed to find k through another method. Please help. I have my cambridge maths exam on 30 may

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    $\begingroup$ Please consider using mathjax $\endgroup$ – Digitalis May 26 '18 at 8:53
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    $\begingroup$ But what is the question in the problem you mention? $\endgroup$ – Bernard May 26 '18 at 8:54
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    $\begingroup$ Good luck for your exam! Also k=-2 $\endgroup$ – Tony Hellmuth May 26 '18 at 8:55
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    $\begingroup$ The first senctence is completely wrong $\endgroup$ – Fakemistake May 26 '18 at 8:58
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Note that

$$-x^2+2x+10=0\iff -\frac12 x^2+x+3=-2$$

then $k=-2$.

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  • $\begingroup$ Precisely the one I was looking for. Thanks $\endgroup$ – Taimur May 26 '18 at 10:16
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$$ 10+2x-x^2=0 $$ $$ x=1-\sqrt{11} \text{ or }x=1+\sqrt{11} $$ $$ y=3+x−0.5x^2 $$ $$ y=3+(1-\sqrt{11})−0.5(1-\sqrt{11})^2 $$ $$ y=-2 $$ $$ y=3+(1+\sqrt{11})−0.5(1+\sqrt{11})^2 $$ $$ y=-2 $$

Therefore since both points lie on $y=-2$, $k=-$2

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