Question

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

Answer 1

`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`

Answer 2

`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"`