Question

if using the method of completing the square to solve the quadratic equation x^2-6x-7=0, which number would have to be added to "complete the square"

Answer 1

We have the next quadratic equation

`x^2-6x-7=0`

in order to have a trinomial perfect square that is

`a^2+2ab+b^2`

in this case, we have

a= x

b= -3

so the expected value

`b^2=(-3)^2=9`

In order to complete the square, we need to add the number 9 on both sides of the equation

`\begin{gathered} x^2-6x-7+9=9 \\ x^2-6x+9=9+7 \\ x^2-6x+9=16 \end{gathered}`
`(x-3)^2=16`

Because we have a quadratic equation we will have 2 solutions for x

`\begin{gathered} (x-3)=\pm\sqrt[]{16} \\ x-3=\pm4 \\ x_1=4+3=7 \\ x_2=-4+3=-1 \end{gathered}`

the solution is

x1=7

x2=-1