SOLUTION: Thanks in advance for all of your help!! Solve by completing the square... 1. {{{x^2}}}+4x+3=0 2. {{{x^2}}}=5x+2 3. 2{{{x^2}}}-8x-9=0 4. 4{{{x^2}}}+2x-3=0

Question 86270: Thanks in advance for all of your help!!
Solve by completing the square...
1. +4x+3=0
2. =5x+2
3. 2-8x-9=0
4. 4+2x-3=0
Answer by scianci(186) (Show Source): You can put this solution on YOUR website!
When completing the square, you want to isolate the constant. For #1, then, you'd have: + 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: = 2
Then, square the result: = 4
Add this to both sides of the equation:
+ 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:
=
x + 1 =
x + 1 =
Solve each equation individually:
x + 1 = 1 , x + 1 = -1
x = 0 , x = -2
Same idea for #2 ; isolate the constant: - 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:
2-8x = 9
Next, factor out the lead coefficient:
2(x^2 - 4x) = 9
Then, complete the square on the - 4x part:
= -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 =
x - 2 = *
x - 2 =
x = 2
#4 proceeds in a similar manner
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