Answer:
B. 3 and 6
Step-by-step explanation:
Given equation:
[tex](x-2)^{\frac{1}{2}}+4=x[/tex]
Subtract 4 from both sides:
[tex]\implies (x-2)^{\frac{1}{2}}+4-4=x-4[/tex]
[tex]\implies (x-2)^{\frac{1}{2}}=x-4[/tex]
Square both sides:
[tex]\implies \left( (x-2)^{\frac{1}{2}}\right)^2=(x-4)^2[/tex]
[tex]\implies x-2=(x-4)^2[/tex]
Expand the brackets on the right side:
[tex]\implies x-2=(x-4)(x-4)[/tex]
[tex]\implies x-2=x^2-8x+16[/tex]
Subtract x from both sides:
[tex]\implies x-2-x=x^2-8x+16-x[/tex]
[tex]\implies -2=x^2-9x+16[/tex]
Add 2 to both sides:
[tex]\implies -2+2=x^2-9x+16+2[/tex]
[tex]\implies 0=x^2-9x+18[/tex]
[tex]\implies x^2-9x+18=0[/tex]
Factor the left side of the equation:
[tex]\implies x^2-6x-3x+18=0[/tex]
[tex]\implies x(x-6)-3(x-6)=0[/tex]
[tex]\implies (x-3)(x-6)=0[/tex]
Apply the zero-product property:
[tex]\implies x-3=0 \implies x=3[/tex]
[tex]\implies x-6=0\implies x=6[/tex]
Therefore, the solution to the equation is:
[tex]x=3, \quad x=6[/tex]
