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Patricia is trying to solve the following equation by completing the square: $$25x^2+20x-10 = 0.$$She successfully rewrites the above equation in the following form: $$(ax + b)^2 = c,$$where $a,$ $b,$ and $c$ are integers and $a > 0.$ What is the value of $a + b + c$?

This is what I tried so far

Given the equation $$25x^2+20x-10=0$$ I first divided everything by $25$ to get $$x^2+\frac{4x}{5}-\frac{2}{5}=0$$ now after completing the square I get $$(x+\frac{2}{5})^2=\frac{14}{25}$$

Now when I completed the square though only $a$ is an integer. What am I missing with the problem?

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    $\begingroup$ Don't divide by $25$ on the first step, and proceed as you would usually. $\endgroup$ – Simply Beautiful Art Mar 7 '17 at 23:13
  • $\begingroup$ Hint: $25=5^2$ so $25x^2+20x=(5x)^2\,+\,2 \cdot(5x) \cdot 2\,+\,\color{red}{?}^{\,2}\,$ $\endgroup$ – dxiv Mar 7 '17 at 23:20
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    $\begingroup$ The question is not well-posed, since if $\,a,b,c\,$ are a solution so too are $\,an,bn,cn^2,\,$ which follows by scaling $\,(ax+b)^2 = c\,$ by $n^2$. That could be remedied in various ways, e.g. by requiring one of the coefficients to be minimal. $\endgroup$ – Bill Dubuque Mar 7 '17 at 23:21
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$(x+\frac{2}{5})^2=\frac{14}{25}$

Can be converted to

$(5x+2)^2=14$

Which would give you $a + b + c = 21$

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If we don't divide by $5$ on the first step, we end up with

$$25x^2+20x+4=14\iff(5x+2)^2=14$$

Thus, $a+b+c=21$.

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