How do you solve #sqrt( x-5) = x-7#?

Question

Eva Abramson · Accepted Answer

#x = {9}# #sqrt(x-5)=x-7# #(sqrt(x-5))^2=(x-7)^2# #x-5=(x-7)^2# #x-5=x^2-14x+49# #x^2-14x-x+49+5=0# #x^2-15x+54=0# #(x-6)(x-9)=0# #if (x-6)=0" "then" "x=6# #if (x-9)=0" "then" "x=9# Watch out for extraneous solutions! 6 does not work in the original equation, however, 9 does. Thus, #{x =9}#

Hazel Anglin · Answer

#x={9}#

square both sides. x-5=(x-7)². x-5=x²-14x+49. 0=x²-15x+54.

#Delta=sqrt((-15)^2-4*1*54)# #Delta=sqrt(225-216)# #Delta=sqrt(9)# #Delta=±3#

#x_1=(15-3)/2#

#x_1=6#

#x_2=(15+3)/2#

#x_2=9# #x={9}#

Eva Adamson · Answer

undefined To solve the equation sqrt(x-5) = x-7, we can follow these steps:

1. Start by isolating the square root term. Square both sides of the equation to eliminate the square root: (sqrt(x-5))^2 = (x-7)^2.

2. Simplify the equation: x-5 = (x-7)^2.

3. Expand the right side of the equation: x-5 = x^2 - 14x + 49.

4. Rearrange the equation to form a quadratic equation: x^2 - 15x + 54 = 0.

5. Solve the quadratic equation by factoring, completing the square, or using the quadratic formula.

6. Once you find the solutions for x, substitute them back into the original equation to check if they satisfy the equation.

7. If the solutions satisfy the equation, they are the valid solutions. If not, there are no solutions to the equation.