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"#