How do you solve #sqrt(9x-8)=x#?

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Answer 1 · 2015-08-10T10:56:30

#{(x = 8), (x=1) :}#

Right from the start, you know that your solution(s) must satisfy two conditions

Overall, you need your solution(s) to satisfy #x>=8/9#.

Start by squaring both sides of the equation to get rid of the radical term

#(sqrt(9x-8))^2 = x^2#

#9x-8 = x^2#

This is equivalent to

#x^2 - 9x + 8 = 0#

The two solutions to this quadratic can be found using the quadratic formula

#x_(1,2) = (-(-9) +- sqrt( (-9)^2 - 4 * 1 * 8))/(2 * 1)#

#x_(1,2) = (9 +- sqrt(49))/2#

#x_(1,2) = (9 +- 7)/2 = {(x_1 = (9 + 7)/2 = 8), (x_2 = (9-7)/2 = 1) :}#

Since both #x_1# and #x_2# satisfy the condition #x>=8/9#, your original equation will have two solutions

#{(x_1 = color(green)(8)), (x_2 = color(green)(1)):}#

Do a quick check to make sure that the calculations are correct

#sqrt(9 * 8 - 8) = 8#

#sqrt(64) = 8 <=> 8 = 8 color(green)(sqrt())#

and

#sqrt(9 * 1 - 8) = 1#

#sqrt(1) = 1 <=> 1 = 1 color(green)(sqrt())#