How do you solve #\sqrt{x^2-5x}-6=0#?

2 Answers
May 20, 2018

Answer:

#x=-4 and x = 9#

Explanation:

#sqrt{x^2-5x}-6=0#

#sqrt{x^2-5x}=6#

#(sqrt{x^2-5x})^2=6^2#

#x^2-5x=36#

#x^2-5x-36 = 0#. factor.

#(x + 4) (x - 9)#

#x=-4 and x = 9#

May 20, 2018

Answer:

#x=-4" or "x=9#

Explanation:

#"isolate "sqrt(x^2-5x)" by adding 6 to both sides"#

#rArrsqrt(x^2-5x)=6#

#color(blue)"square both sides"#

#(sqrt(x^2-5x))^2=6^#

#rArrx^2-5x=36#

#"rearrange into "color(blue)"standard form";ax^2+bx+c=0#

#"subtract 36 from both sides"#

#rArrx^2-5x-36=0larrcolor(blue)"in standard form"#

#"the factors of - 36 which sum to - 5 are - 9 and + 4"#

#rArr(x-9)(x+4)=0#

#"equate each factor to zero and solve for x"#

#x+4=0rArrx=-4#

#x-9=0rArrx=9#

#color(blue)"As a check"#

Substitute these values into the left side of the equation and if equal to the right side then they are the solutions.

#x=-4tosqrt(16+20)-6=sqrt36-6=6-6=0#

#x=9tosqrt(81-45)-6=sqrt36-6=6-6=0#

#rArrx=-4" or "x=9" are the solutions"#