EXPLANATION:
Given;
We are given the quadratic equation as shown below;
[tex]x^2-8x+13=0[/tex]
Required;
We are required to solve for x by completing the square method.
Step-by-step solution;
We start with the constant 13.
Subtract 13 from both sides of the equation;
[tex]x^2-8x+13-13=0-13[/tex][tex]x^2-8x=-13[/tex]
Next we take the coefficient of x (that is -8). We half it, and then square the result. After that we add it to both sides of the equation;
[tex]\begin{gathered} \frac{1}{2}\times-8=-\frac{8}{2} \\ Next: \\ (-\frac{8}{2})^2 \end{gathered}[/tex]
We now have;
[tex]x^2-8x+(-\frac{8}{2})^2=-13+(-\frac{8}{2})^2[/tex]
We can now simplify this;
[tex]x^2-8x+(-4)^2=-13+(-4)^2[/tex][tex]x^2-8x+16=-13+16[/tex][tex]x^2-8x+16=3[/tex]
We now factorize the left side of the equation;
[tex]\begin{gathered} x^2-4x-4x+16 \\ (x^2-4x)-(4x-16) \\ x(x-4)-4(x-4) \\ (x-4)(x-4) \\ Therefore: \\ (x-4)^2 \end{gathered}[/tex]
we can now refine our equation to become;
[tex](x-4)^2=3[/tex]
We can now solve for x as follows;
Take the square root of both sides;
[tex]x-4=\pm\sqrt{3}[/tex]
Therefore;
[tex]\begin{gathered} x-4=\sqrt{3} \\ x=\sqrt{3}+4 \\ Also: \\ x-4=-\sqrt{3} \\ x=-\sqrt{3}+4 \end{gathered}[/tex]
ANSWER:
[tex]\begin{gathered} x_1=4+\sqrt{3} \\ x_2=4-\sqrt{3} \end{gathered}[/tex]
