How do you solve #5 sqrt (x-6) = x#?

1 Answer
Jul 25, 2016

Answer:

x = 10 , x = 15

Explanation:

Begin by dividing both sides of the equation by 5.

#(cancel(5)^1sqrt(x-6))/cancel(5)^1=x/5#

#rArrsqrt(x-6)=x/5#

To obtain the value inside the square root, we #color(blue)"square both sides"#

#color(orange)"Reminder" color(red)(|bar(ul(color(white)(a/a)color(black)(sqrtaxxsqrta=(sqrta)^2=a)color(white)(a/a)|)))#

#color(blue)"squaring both sides"#

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

#rArrx-6=x^2/25#

Now multiply both sides by 25 to eliminate the fraction.

#rArr25(x-6)=cancel(25)^1xx(x^2)/cancel(25)^1rArr25(x-6)=x^2#

#color(orange)"Reminder"#

The standard form of a quadratic equation is.

#color(red)(|bar(ul(color(white)(a/a)color(black)(ax^2+bx+c=0)color(white)(a/a)|)))#

Rearrange # 25(x-6)=x^2" into standard form"#

#rArr25x-150=x^2rArrx^2-25x+150=0#

To factorise we consider the product of the factors of 150 which also sum to -25 . These are -10 and -15

#rArr(x-10)(x-15)=0" and solving gives"#

#x=10,x=15#