How do you solve #sqrt[7x+2]-2=7x# and find any extraneous solutions?

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Answer 1 · 2018-06-15T04:49:25

#x = -1/7# or #x = -2/7#

Move the constant term

#sqrt(7x+2) = 7x + 2#

Square both sides

#(sqrt(7x+2)) ^ 2 = (7x + 2) ^ 2#

#7x+2 = 49x^2 + 28x + 4#

Move everything over

#0 = 49x ^2 +21x+2#

#x_(1,2) = (-b ± sqrt(b ^2 - 4ac)) / (2a) #

So

#x_(1, 2) = (-21 ± sqrt(21 ^2 - 4(49)(2))) / (2(49)) #

Solve to get

#x = -1 / 7" "# or #" "x = -2 / 7#

To check for extraneous roots, plug the solutions back into the original equation.

#sqrt(7(-1/7)+2) = 7(-1/7) + 2#

#1 = 1 -># the solution #(-1/7)# is correct!

Repeat for the other solution

#sqrt(7(-2/7)+2) = 7(-2/7) + 2#

#0 = 0 -># the solution #(-2/7)# is also correct!

Extraneous solutions appear when your solutions don't satisfy the original equation.