How do you solve #sqrt( 7x^2-1) + 1=3x#?

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Answer 1 · 2015-07-24T19:02:44

Rewrite this equation as a quadratic and solve for #x#.

The first thing you need to do in order to solve this equation is isolate the radical on one side of the equation.

To do that, add #-1# to both sides of the equation

#sqrt(7x""^2 - 1) + cancel(1) - cancel(1) = 3x - 1#

#sqrt(7x""^2 - 1) = 3x-1#

Next, square both sides of the equation to get rid of the radical

#(sqrt(7x""^2 - 1))^2 = (3x-1)^2#

#7x""^2 - 1 = 9x""^2 - 6x + 1#

Move all the terms on one side of the equation

#2x""^2 -6x +2 = 0#

You can use the quadratic formula to determine the two solutions

#x_(1,2) = (6 +- sqrt(36 - 4 * 2 * 2))/4#

#x_(1,2) = (6 +- sqrt(20))/4 = (3 +- sqrt(5))/2 => { (x_1 = (3 + sqrt(5))/2), (x_2 = (3 - sqrt(5))/2) :}#

You need to check these two solutions to see if both are valid, i.e. if you don't have an extraneous solution.

For #x_1# you have

#sqrt(7 * ((3 + sqrt(5))/2)^2-1) = 3 * ((3 + sqrt(5))/2) -1#

This is equivalent to

#sqrt(94 + 42sqrt(5))/cancel(2) = (7 + 3sqrt(5))/cancel(2)#

Square both sides of the equation to get

#(sqrt(94 + 42sqrt(5)))^2 = (7 + 3sqrt(5))^2#

#94 + 42sqrt(5) = 49 + 42sqrt(5) + 45# #-># #x_1# is #color(green)("valid")#

For #x_2# you have

#sqrt(7 * ((3 - sqrt(5))/2)^2-1) = 3 * ((3 - sqrt(5))/2) -1#

This is equivalent to

#94 - 42sqrt(5)) = 49 - 42sqrt(5) + 45# #-># #x_2# is #color(green)("valid")#