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How do I complete the square when the second degree coefficient is greater than one. I can do it when $x^2+4x-4=0$, for example, but I can't work out how to do when $3x^2+4x-4=0$.

Alice Ryhl
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Jeremy
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4 Answers4

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$$3x^2+4x-2=3\left(x^2+\frac43x-\frac43\right)=3\left[\left(x+\frac23\right)^2-\frac49-\frac43\right]=\ldots$$

DonAntonio
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Hint: $$3x^2+4x-4 = 3\left(x^2+\frac{4}{3}x-\frac{4}{3}\right).$$

Michael Albanese
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Best solution is to normalize the leading coefficient. I.E. $$3x^2+4x-4=0 \implies x^2 +\frac{4x}{3}-\frac{4}{3}=0 \implies (x^2+\frac{4x}{3}+ ~~~)-\frac{4}{3}=0$$ We know that for completing the square, we should have the $c$ term equal to the square of half the $b$ term divided by the $a$ term. $c=\left(\frac{b}{2a}\right)^2$ $$ (x^2+\frac{4x}{3}+ \frac{4}{9})-\frac{4}{9}-\frac{4}{3}=0 \implies \left(x+\frac{2}{3}\right)^2-\frac{16}{9}=0$$ Solve: $$\left(x+\frac{2}{3}\right)^2=\frac{16}{9} \implies \left( x+\frac{2}{3}\right)=\pm\frac{4}{3}$$ Therfore: $$x=\frac{2}{3},-2$$

FundThmCalculus
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  • How can I do it without normalizing the leading coefficient? – Jeremy May 29 '14 at 12:23
  • It's not particularly easy to do that way, since then you would have radicals on the leading term (sqrt(3)*x+[constant])^2, which would get messy in a real hurry. – FundThmCalculus May 29 '14 at 12:32
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If you would prefer to avoid fractions as long as possible multiply by four times the leading coefficient:

$$12(3x^2+4x-4)=36x^2+48x-48=(6x+4)^2-64$$ In general $$4a(ax^2+bx+c)=4a^2x^2+4abx+4ac=(2ax+b)^2+4ac-b^2$$

Mark Bennet
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