How do you solve \sqrt{x^2-5x}-6=0?
2 Answers
Explanation:
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"
