Question

Solve x2 + 6x = 7 by completing the square. Which is the solution set of the equation?

Answer 1

A word to the wise:  Please use " ^ " to indicate exponentiation:  x^2, not 2x.

x^2 + 6x = 7   can be re-written as   x^2 + 6x                               = 7

Take HALF of the coefficient of x; in other words, take HALF of 6.  Result:  3.

Square this result   Add the square to x^2 + 6x, and then subtract it from
                                                             x^2 + 6x + 9:    x^2 + 6x + 9     -9  = 7

Now rewrite x^2 + 6x + 9 in the form (x+3)^2  Equate that to 7+9 (or 16) on the right side.

Then you have (x+3)^2 = 16.

Take the square root of both sides:    x+3 = sqrt(16) = plus or minus 4.

There are 2 roots:  One is x = -3 + 4, or   x = 1.

The other is                       x = -3 - 4, or x = -7

It's very important that you check your results.    Is x =1 a solution?

Subst. 1 for x   in   x^2 + 6x = 7:   1^2 + 6(1) = 7?  Yes, this is true, so x = 1 is a solution.

Subst. -7 for x in   x^2 + 6x = 7:  (-7)^2 + 6(-7) = 7     Is this true?

49 - 42 = 7?  Yes, this is true, so x = -7 is another solution.

The "solution set" is {1, -7}.