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.

Related questions
See all questions in Radical Equations
Impact of this question
1243 views around the world
You can reuse this answer
Creative Commons License