How to solvE x^2 + 6x – 7 = 0 by completing the square.
I'll do the same procedure as in the first exercise, in exactly the same order. (Study tip: Always working these problems in exactly the same way will help you remember the steps when you're taking your tests.)
First, I write down the equation they've given me.
x^2 + 6x – 7 = 0
I move the constant term (the loose number) over to the other side of the "equals".
x^2 + 6x = 7
The leading term is already only multiplied by 1, so I don't have to divide through by anything. So that step is done.
Now I'll grab some scratch paper, and do my computations. First, the coefficient of the "linear" term (that is, the term with just x, not the x^2 term), with its sign, is:
Numerical coefficient: +6
I'll multiply this by 1/2
derived value: (+6)(1/2)=3
My next step is to square this derived value:
square of derived value: (3)^2 = 9
Now I go back to my equation, and add this squared value to either side:
x^2 + 6x + 9 = 7 + 9
I'll simplify the strictly-numerical stuff on the right-hand side:
x^2 + 6x + 9 = 16
And now I'll convert the left-hand side to completed-square form, using the derived value (which I circled in my scratch-work, so I wouldn't lose track of it), along with its sign:
(x + 3)^2 = 16
Now that the left-hand side is in completed-square form, I can square-root each side, remembering to put a "plus-minus" on the strictly-numerical side:
...and then I'll solve for my two solutions:
x = –3 ±4
= –3 – 4, –3 + 4
= –7, 1
Then my answer is:
x = –7, 1