How do you complete the square when a quadratic equation has a coefficient?

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How do you complete the square when a quadratic equation has a coefficient?

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Answer 1 · 2015-02-20T12:36:35

It is easier to explain with an example:

$3 x^{2} + 7 x - 5 = 0 \Rightarrow 3 (x^{2} + \frac{7}{3} x) - 5 = 0$ $3x^2+7x-5=0rArr3(x^2+7/3x)-5=0$

Now, since $\frac{7}{3} x$ $7/3x$ will have to be the double product of the first and the second terms of the square, if we divide it for the double of the first ( $2 x$ $2x$ ) we can obtain the second:

$\frac{\frac{7}{3} x}{2 x} = \frac{7}{6}$ $(7/3x)/(2x)=7/6$ , so:

$3 (x^{2} + \frac{7}{3} x + {(\frac{7}{6})}^{2} - {(\frac{7}{6})}^{2}) - 5 = 0 \Rightarrow$ $3(x^2+7/3x+(7/6)^2-(7/6)^2)-5=0rArr$

$3 (x^{2} + \frac{7}{3} x + \frac{49}{36} - \frac{49}{36}) - 5 = 0 \Rightarrow$ $3(x^2+7/3x+49/36-49/36)-5=0rArr$

$3 (x^{2} + \frac{7}{3} x + \frac{49}{36}) - 3 \cdot \frac{49}{36} - 5 = 0 \Rightarrow$ $3(x^2+7/3x+49/36)-3*49/36-5=0rArr$

$3 {(x + \frac{7}{6})}^{2} - \frac{49}{12} - 5 = 0 \Rightarrow$ $3(x+7/6)^2-49/12-5=0rArr$

$3 {(x + \frac{7}{6})}^{2} = \frac{49 + 60}{12} \Rightarrow$ $3(x+7/6)^2=(49+60)/12rArr$

$3 {(x + \frac{7}{6})}^{2} = \frac{109}{12} \Rightarrow$ $3(x+7/6)^2=109/12rArr$ and, if you want to solve:

${(x + \frac{7}{6})}^{2} = \frac{109}{36} \Rightarrow (x + \frac{7}{6}) = \pm \sqrt{\frac{109}{36}} \Rightarrow$ $(x+7/6)^2=109/36rArr(x+7/6)=+-sqrt(109/36)rArr$

$x = - \frac{7}{6} \pm \frac{\sqrt{109}}{6} = \frac{- 7 \pm \sqrt{109}}{6}$ $x=-7/6+-sqrt109/6=(-7+-sqrt109)/6$ .

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How do you complete the square when a quadratic equation has a coefficient?

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