I posted this q yesterday but got no answer.

Can you move all of the radicals in the quadratic formula to the denominator? Please show your steps so I can understand.

Guest Mar 15, 2017

#1**0 **

It's possible.....

Working with one of the solutions, we have

[ -b + sqrt ( b^2 - 4ac) ] / (2a)

Multiply the numerator and the denominator by -b - sqrt[b^2 - 4ac]....this gives us

[ -b + sqrt ( b^2 - 4ac) ] [ - b - sqrt (b^2 - 4ac) ] / [ (2a) (-b - sqrt(b^2 - 4ac) ) ] =

[ b^2 - (b^2 - 4ac) ] / [ (2a) (-b - sqrt(b^2 - 4ac) ) ] =

[ 4ac] / [ (2a) (-b - sqrt(b^2 - 4ac) ) ] =

- [ 2c } / [ b + sqrt(b^2 - 4ac) ]

We could do a similar thing with the other solution, [ -b - sqrt ( b^2 - 4ac) ] / (2a)

CPhill
Mar 15, 2017