How do you solve #sqrt(2x+1/6)=x+5/6#?

1 Answer
Tony B
Mar 21, 2017

I have taken you to a point where you can finish it off.

Explanation:

If you have a square root on one side of the equals and a 'non' square root on the other, you can 'get rid' of the root by squaring both sides. Giving:

#2x+1/6=(x+5/6)^2#

#2x+1/6=x^2+5/3x+25/36#

Collecting terms on one side only of the equals to give us a quadratic equation.

#x^2+5/3x-2x+25/36-1/6=0#

#x^2-1/3x +19/36=0#

Compare to #y=ax^2+bx+c# where #x=(-b+-sqrt(b^2-4ac))/(2a)#

#=> x=(1/3+-sqrt((1/9)-4(1)(19/36)))/(2(1))#

I will let you finish this off.

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