Question 86270
When completing the square, you want to isolate the constant. For #1, then, you'd have: {{{x^2}}}+ 4x = -3.
The next step, if applicable, is to factor out the lead coefficient [Not necessary in this case ; I'll demonstrate it in solving #3]
Then, divide the linear coefficient, in this case 4, by 2: {{{4/2}}} = 2
Then, square the result: {{{2^2}}} = 4
Add this to both sides of the equation:
{{{x^2}}}+ 4x + 4 = -3 + 4
Now, factor the left side. It is a perfect-square trinomial which is what completing the square creates, so it'll factor as the square of a binomial:
(x + 2)^2 = 1
Take square roots of both sides:
{{{sqrt(x + 1)^2)}}} = {{{(+-sqrt(1))}}}
x + 1 = {{{(+-sqrt(1))}}}
x + 1 = {{{+-1}}}
Solve each equation individually:
x + 1 = 1 , x + 1 = -1
x = 0 , x = -2
Same idea for #2 ; isolate the constant: {{{x^2}}} - 5x = 2
Then proceed exactly as in #1 above. I'll let you finish it.
For #3, start off the same way, by isolating the constant:
2{{{x^2}}}-8x = 9
Next, factor out the lead coefficient:
2(x^2 - 4x) = 9
Then, complete the square on the {{{x^2}}} - 4x part:
{{{-4/2}}} = -2 ; {{{(-2)^2}}} = 4
2(x^2 - 4x + 4) = 9 + 2(4) [don't forget the 2 that got factored out at the start]
2(x - 2)^2 = 17
(x - 2)^2 = 17/2
x - 2 = {{{+-sqrt(17/2)}}}
x - 2 = {{{+-sqrt(17/2)}}}*{{{sqrt(2)/sqrt(2)}}}
x - 2 = {{{(+-sqrt(34))/2}}}
x = 2{{{(+-sqrt(34))/2}}}
#4 proceeds in a similar manner