Given the equation:
[tex]\mleft(x-2\mright?^{})^{\frac{1}{2}}+4=x[/tex]1. You need to remember the following property:
[tex]\sqrt[n]{b^m}=b^{\frac{m}{n}}[/tex]2. Then, you can rewrite the equation as follows:
[tex]\sqrt[]{x-2}^{}+4=x[/tex]3. Now you need to apply the Subtraction Property of Equality by subtracting 4 from both sides of the equation:
[tex]\begin{gathered} \sqrt[]{x-2}^{}+4-(4)=x-(4) \\ \\ \sqrt[]{x-2}^{}=x-4 \end{gathered}[/tex]4. Square both sides of the equation:
[tex]\begin{gathered} (\sqrt[]{x-2})^{}=(x-4)^2 \\ x-2=(x-4)^2 \end{gathered}[/tex]5. Apply the following on the right side:
[tex](a-b)^2=a^2-2ab+b^2[/tex]Then:
[tex]\begin{gathered} x-2=(x^2-2(x)(4)+4^2) \\ \\ x-2=x^2-8x+16 \end{gathered}[/tex]6. Write the Quadratic Equation in this form:
[tex]ax^2+bx+c=0[/tex]Then:
[tex]\begin{gathered} 0=x^2-8x+16-x+2 \\ x^2-9x+18=0 \end{gathered}[/tex]7. In order to factor it, you can find two numbers whose sum is -9 and whose product is 18. These numbers would be -3 and -6. Then:
[tex](x-3)(x-6)=0[/tex]8. Notice that you get these values of "x":
[tex]\begin{gathered} x-3=0\Rightarrow x_1=3 \\ \\ x-6=0\Rightarrow x_2=6_{} \end{gathered}[/tex]9. Substitute each value of "x" into the equation and evaluate, in order to check if they are solutions to the equation:
- For:
[tex]x_1=3[/tex]You get:
[tex]\begin{gathered} \sqrt[]{(3)-2}^{}+4=(3) \\ \sqrt[]{1}^{}+4=3 \\ 5=3\text{ (False)} \end{gathered}[/tex]It is not a solution.
- For:
[tex]x_2=6[/tex]You get:
[tex]\begin{gathered} \sqrt[]{(6)-2}^{}+4=(6) \\ \sqrt[]{4}+4=6 \\ 2+4=6 \\ 6=6\text{ (True)} \end{gathered}[/tex]It is a solution.
Hence, the answer is:
