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If $4$ is one solution of quadratic equation $x^2 + 3x +k = 10$ What is the other solution? I know how to solve a quadratic equation but how do I solve this ? Answer is $-7$

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    Solution *where*? Quadratics can have more than two solutions, e.g. $\rm\:x^2 = 1\:$ has $4$ solutions $\rm\:\pm1,\pm3\:$ in $\:\Bbb Z/8 = $ integers mod $8.\ \ $2012-08-29

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Hint: You should use the solution you are given to determine $k$, then solve the resulting equation.

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    So is the solution k ?2012-08-29
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    No. Judging on the extremely fast (several seconds) turnaround in your answer to me, I'm sure you need to think more about it. Don't guess: be systematic!2012-08-29
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    Just look at what you're given: "4 is a solution". If you use that hypothesis *at all* then you get $k$ right away.2012-08-29
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    Figured out what you meant. Thanks2012-08-29
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We don't ned to find $k$. The sum of the solutions is $-3$. If one solution is $4$, the other must be $-7$.

Remark: If $ax^2+bx+c=0$ is a quadratic equation, then the sum of the roots is $-\frac{b}{a}$, and the product of the roots is $\frac{c}{a}$. These are important and often-used results.

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    Mentioning Vieta's formula for $n=2$ might be a good idea.2012-08-29
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    Judging from the OP, I think this is a little too slick. Needs more explanation. As it is now, it isn't much better than typing "-7".2012-08-29
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If the solutions of $x^2+ax+b=0$ are $x=c$ and $x=d$, then $x^2+ax+b$ factors as $$x^2+ax+b=(x-c)(x-d)$$ Multiplying out the right side you get $$x^2+ax+b=x^2-(c+d)x+cd$$ so that tells you $$a=-(c+d),\qquad b=cd$$ Now in your situation you know $a$ and you know one of the roots, $c$; can you work out $d$ from the above?

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    Yep thanks. Just need to use $a=-(c+d)$ so $3=-4-d$ or d=-7 where c and d are the two solutions2012-08-29