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:

$3x^2+7x-5=0rArr3(x^2+7/3x)-5=0$

Now, since $7/3x$ 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 ( $2x$ ) we can obtain the second:

$(7/3x)/(2x)=7/6$ , so:

$3(x^2+7/3x+(7/6)^2-(7/6)^2)-5=0rArr$

$3(x^2+7/3x+49/36-49/36)-5=0rArr$

$3(x^2+7/3x+49/36)-3*49/36-5=0rArr$

$3(x+7/6)^2-49/12-5=0rArr$

$3(x+7/6)^2=(49+60)/12rArr$

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

$(x+7/6)^2=109/36rArr(x+7/6)=+-sqrt(109/36)rArr$

$x=-7/6+-sqrt109/6=(-7+-sqrt109)/6$ .

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

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