Part a. We are given the following quadratic equation:
[tex]x^2-6x=-5[/tex]This is an equation of the form:
[tex]x^2+bx=c[/tex]To complete the square we will add and subtract the following term:
[tex](\frac{b}{2})^2[/tex]Substituting we get:
[tex](\frac{-6}{2})^2[/tex]Solving the operations and simplifying we get:
[tex](\frac{-6}{2})^2=(-3)^2=9[/tex]Therefore, the value of "c" is 9.
Part b. We will substitute the value of "c" in the equation:
[tex]x^2-6x+9=-5+9[/tex]Now, we solve the operation on the right side:
[tex]x^2-6x+9=4[/tex]Now, we will factor the left side using the square of a binomial. Therefore, we take the square root of the first and third term and rearrange them in the form of the square of a binomial, like this:
[tex](x-3)^2=4[/tex]This completes part B.
Part C. Now, we will solve for "x". To do that we will take the square root to both sides:
[tex]\begin{gathered} x-3=\sqrt[]{4} \\ x-3=\pm2 \end{gathered}[/tex]Now we add 3 to both sides:
[tex]x=3\pm2[/tex]Since we have a quadratic equation there are two possible solutions for "x". The first solution is determined using the plus sign:
[tex]x=3+2=5[/tex]The second solution is determined using the minus sign:
[tex]x=3-2=1[/tex]Therefore, the values of "x" are 5 and 1.
