How do you solve #5 sqrt (x-6) = x#?
1 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#