Re: Quadratic Equation help

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

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

multiply through by 4*2 = 8

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

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

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

The square has been completed on the left.

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

Take square root

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

add 3

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

divide by 4

$\displaystyle 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.

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

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

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

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

a = c^2 - 1

b = -1

c = -1

plug it in

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

Simplify.

:)

Re: Quadratic Equation help

Re: Quadratic Equation help

Quote:

Originally Posted by

**Caddym89** Cheers agentmulder!

I'll drink to that!

:)

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Re: Quadratic Equation help

Re: Quadratic Equation help

Quote:

Originally Posted by

**ibdutt**

You can complete the square this way too...

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

$\displaystyle ax^2 + bx = -c $

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

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

add b^2 to both sides

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

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

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

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

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

:)

Re: Quadratic Equation help

Quote:

Originally Posted by

**Caddym89** Hi there, I'm having trouble with two of the more challenging problems in my homework.

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

2) c^{2}x^{2 }- x - 1 = x^{2}, 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.

$\displaystyle \displaystyle \begin{align*} 2x^2 - 3x + 2a &= 0 \\ x^2 - \frac{3}{2} x + a &= 0 \\ x^2 - \frac{3}{2}x + \left( -\frac{3}{4} \right) ^2 - \left( -\frac{3}{4} \right) ^2 + a &= 0 \\ \left( x - \frac{3}{4} \right) ^2 - \frac{9}{16} + \frac{16a}{16} &= 0 \\ \left( x - \frac{3}{4} \right) ^2 &= \frac{9 - 16a}{16} \\ x - \frac{3}{4} &= \frac{\pm \sqrt{ 9 - 16a }}{4} \\ x &= \frac{3 \pm \sqrt{ 9 - 16a }}{4} \end{align*}$

Re: Quadratic Equation help

Thanks everyone, it's useful to know more than one method of completing the square also. My problem is making silly mistakes, I was originally stuck on this trying to figure out how to get the 16a in prove it's method, this is again not so difficult. (;