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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
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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)  

 

 

 

cool cool cool

CPhill  Mar 15, 2017

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