Question

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

Answer 1

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`.