• Apr 9th 2013, 09:57 PM
Hi there, I'm having trouble with two of the more challenging problems in my homework.

1) 2x2 - 3x + 2a = 0 , this needs to be solved using complete the square method.

2) c2x2 - x - 1 = x2, to be solved using quadratic formula.

any help would be appreciated, thanks.
• Apr 9th 2013, 11:21 PM
agentmulder
1) To complete a square isolate the constant term, multiply everything by 4 times the coeficient of x^2, square the coefficient of x and add it to both sides. It looks like this

$2x^2 - 3x = -2a$

multiply through by 4*2 = 8

$16x^2 - 24x = -16a$

add (-3)^2 = 9 to both sides

$16x^2 - 24x + 9 = 9 - 16a$

The square has been completed on the left.

$(4x - 3)^2 = 9 - 16a$

Take square root

$4x - 3 = \pm \sqrt{9 - 16a}$

$4x = 3 \pm \sqrt{9 - 16a}$

divide by 4

$x = \frac{3 \pm \sqrt{9 - 16a} }{4}$

2) To use the popular version of the quadratic formula set the equation to 0 then identify a, b, and c then plug into the formula.

$c^2x^2 - x^2 - x - 1 = 0$

$(c^2 - 1)x^2 - x - 1 = 0$

$If \ ax^2 + bx + c = 0$

$Then \ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

a = c^2 - 1
b = -1
c = -1

plug it in

$x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(c^2 - 1)(-1)}}{2(c^2 - 1)}$

Simplify.

:)
• Apr 10th 2013, 12:48 AM
Cheers agentmulder!
• Apr 10th 2013, 12:54 AM
agentmulder
Quote:

Cheers agentmulder!

I'll drink to that!

:)
• Apr 10th 2013, 04:16 AM
ibdutt
• Apr 10th 2013, 05:02 AM
agentmulder
Quote:

Originally Posted by ibdutt

You can complete the square this way too...

$ax^2 + bx + c = 0$

$ax^2 + bx = -c$

$4a(ax^2 + bx) = -4ac$

$4a^2x^2 + 4abx = -4ac$

$4a^2x^2 + 4abx + b^2 = b^2 - 4ac$

$(2ax + b)^2 = b^2 - 4ac$

$2ax + b = \pm \sqrt{b^2 - 4ac}$

$2ax = -b \pm \sqrt{b^2 - 4ac}$

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

:)
• Apr 10th 2013, 05:48 PM
Prove It
Quote:

Hi there, I'm having trouble with two of the more challenging problems in my homework.

1) 2x2 - 3x + 2a = 0 , this needs to be solved using complete the square method.

2) c2x2 - x - 1 = x2, to be solved using quadratic formula.

any help would be appreciated, thanks.

While I do like Agent Mulder's version of completing the square, the more common method is to take out whatever the a term is as a factor first and complete the square on the remaining stuff.

\displaystyle \begin{align*} 2x^2 - 3x + 2a &= 0 \\ x^2 - \frac{3}{2} x + a &= 0 \\ x^2 - \frac{3}{2}x + \left( -\frac{3}{4} \right) ^2 - \left( -\frac{3}{4} \right) ^2 + a &= 0 \\ \left( x - \frac{3}{4} \right) ^2 - \frac{9}{16} + \frac{16a}{16} &= 0 \\ \left( x - \frac{3}{4} \right) ^2 &= \frac{9 - 16a}{16} \\ x - \frac{3}{4} &= \frac{\pm \sqrt{ 9 - 16a }}{4} \\ x &= \frac{3 \pm \sqrt{ 9 - 16a }}{4} \end{align*}
• Apr 10th 2013, 06:57 PM