# Solution to quadratic equation from an earlier solution

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.\ \$ –  Bill Dubuque Aug 29 '12 at 4:22

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 solutions –  MistyD Aug 29 '12 at 1:50

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 ? –  MistyD Aug 29 '12 at 1:43
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! –  rschwieb Aug 29 '12 at 1:44
Just look at what you're given: "4 is a solution". If you use that hypothesis at all then you get $k$ right away. –  rschwieb Aug 29 '12 at 1:49
Figured out what you meant. Thanks –  MistyD Aug 29 '12 at 1:53
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.
Mentioning Vieta's formula for $n=2$ might be a good idea. –  Pedro Tamaroff Aug 29 '12 at 1:46