Solving quadratic using complete the square method

[ryana]

Solve: 2x^2 + 3x - 2 = 0

1. Preserve the constant
2x^2 + 3x = 2

2. Square 1/2 the 3 - the constant of 3x
3/2 * 3/2 = 9/4

3. Add the value to both sides
2x^2 + 3x + 9/4 = 2 + 9/4

4. Multiply by 4 to eliminate the denominator
8x^2 + 12x + 9 = 17

Now the constant is a square but the first term is no longer a square and I find myself stuck.

I tried to move the 17 and divide by 2 so that I get two squares but -4 is an impossible square
8x^2 + 12x -8 = 0
4x^2 + 6x - 4 = 0

 

[Loren]

Solve: 2x^2 + 3x - 2 = 0
I always advocate getting the coefficient of the x^2 term equal to one. In your case, divide both sides of the equation by 2. Then do the steps that you have indicated.

 

[jwpaine]

Solve: 2x^2 + 3x - 2 = 0


First thing is first: to complete the square, you need to be doing ((1/2)b)^2 AFTER you have pull out / divide by your "a" in ax^2.
Second, you are making it too difficult by multiplying stuff and then having ANOTHER constant in front of your terms that you have to deal with.

(1) If you want to get it in the form of a(x - h)^2 + k, you can pull out the 2:

\(\displaystyle 2(x^2 + \frac{3}{2}x) = 2\)
\(\displaystyle 2(x^2 + \frac{3}{2}x + (\frac{1}{2} \cdot \frac{3}{2})^2 = 2 + 2( (\frac{1}{2} \cdot \frac{3}{2})^2 )\)
\(\displaystyle 2(x + \frac{1}{2} \cdot \frac{3}{2})^2 = 2 + 2( (\frac{1}{2} \cdot \frac{3}{2})^2 )\)

Or...
(2) If you simply want the roots, right off, you could just divide everything out by 2 so you have \(\displaystyle x^2 + \frac{3}{2}x + 1 = 0\) and then proceed to "complete the square", square root both sides, and solve for x

A good practice for completing the square is deriving the popular "quadratic formula"
With ax^2 + bx + c = 0, derive the quadratic formula x = [-b +/- sqrt[b^2 - 4(a)(c)]] / 2a if you can do that, than you have learned the method of completing the square, correctly.

Cheers!
John

 

[ryana]

I can't follow the squaring very well but the formula works pretty well.

 

[tkhunny]

You just have to stare at it until it soaks in.

(a+b)^2 = a^2 + 2ab + b^2

Given this: a^2 + 2ab

How do you get the third term?

2ab <== Get rid of the 'a'. That's the relative of the first term

2b <== Half of that

b <== Squared

b^2 <== Staring, Soaking, Staring and Soaking...