I think that I did this one correctly as well but I could not figure out the factoring.

√x+2 + √3x+7=1

√3x+7 = 1-√x+2

√(3x+7)^2 =(1-√x+2)^2

3x+7=1-x = 2-2√x+2

3x+7 = -x+3-2√x+2

2x +4 = (-2√x+2)

(2x+4)^2= (2√x+2)^2

4x^2+16x+16=4(x+2)

4x^2+12x-8=0

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- Feb 18th 2009, 08:58 PMroachbeatsSolving Radical expressions
I think that I did this one correctly as well but I could not figure out the factoring.

√x+2 + √3x+7=1

√3x+7 = 1-√x+2

√(3x+7)^2 =(1-√x+2)^2

3x+7=1-x = 2-2√x+2

3x+7 = -x+3-2√x+2

2x +4 = (-2√x+2)

(2x+4)^2= (2√x+2)^2

4x^2+16x+16=4(x+2)

4x^2+12x-8=0 - Feb 18th 2009, 10:35 PMearboth