Hi I came across the following question where I need to find $$mk$$ from $$ (x-2) (x+k) = x^2 + mx - 10 $$ The answer is 15. Any suggestions on how I could do that ?
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The sum of the roots $(2, -k)$ equals $-m$. The product of the roots $-2k=-10$. Therefore $k=5$ and $m=3$. |
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You can find the product $(x-2)(x+k)$, getting $x^2+(k-2)x-2k$. This is supposed to be the same polynomial as $x^2+mx-10$. So the constant terms must match, and the coefficients of $x$ must match. That gives us $-2k=-10$ amd $k-2=m$. From $-2k=10$, we conclude that $k=5$. Then from $k-2=m$ we conclude that $m=3$. It follows that $mk=15$. |
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Expression$$ (x-2) (x+k) = x^2 + mx - 10 $$ can be rewriten as $$ x^2+(k-2)x-2k = x^2 + mx - 10 $$ equating the coefficients next to same power of $x$ we get that $k-2=m$ and $-2k=-10$ or$ k=5$ and $m=5-2=3$ that means $$mk=5\times 3=15$$ |
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