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

Question

Abigail Allen · Accepted Answer

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

#"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"#

Kennedy Adams · Answer

#x=-4 and x = 9# #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#

Julia Adams · Answer

undefined To solve the equation \sqrt{x^2-5x}-6=0, we can isolate the square root term and then square both sides of the equation. Here are the steps:

1. Add 6 to both sides of the equation: \sqrt{x^2-5x} = 6.
2. Square both sides of the equation: (\sqrt{x^2-5x})^2 = 6^2.
3. Simplify: x^2-5x = 36.
4. Rearrange the equation to form a quadratic equation: x^2-5x-36 = 0.
5. Factor the quadratic equation: (x-9)(x+4) = 0.
6. Set each factor equal to zero and solve for x: x-9 = 0 or x+4 = 0.
7. Solve for x: x = 9 or x = -4.

Therefore, the solutions to the equation \sqrt{x^2-5x}-6=0 are x = 9 and x = -4.