Difference between revisions of "Quadratic equation"

Revision as of 20:59, 17 June 2006

Example: Solve the equation $x^2-3x+2=0$ for x. Solution: $x^2-3x+2=0$ First we expand the middle term. This is different for all quadratics. We cleverly choose this so that it has common factors. We now have $x^2-x-2x+2=0$ Next, we factor out our common terms to get: $x(x-1)-2(x-1)=0$ . We can now factor the (x-1) term to get: $(x-1)(x-2)=0$ . By a well know theorem, Either $(x-1)$ or $(x-2)$ equals zero. We now have the pair or equations x-1=0, or x-2=0. These give us answers of x=1 or x-2. Plugging these back into the original equation, we find that both of these work! We are done.

Completing the square

Quadratic Formula

See Quadratic Formula.

@@ Line 17: / Line 17: @@
 === Quadratic Formula ===
-===General Solution For A Quadratic by Completing the Square===
+See [[Quadratic Formula]].
-Let the quadratic be in the form <math>a\cdot x^2+b\cdot x+c=0</math>.
-Moving c to the other side, we obtain
-<math>ax^2+bx=-c</math>
-Dividing by <math>{a}</math> and adding <math>\frac{b^2}{4a^2}</math> to both sides yields
-<math>x^2+\frac{b}{a}x+\frac{b^2}{4a^2}=-\frac{c}{a}+\frac{b^2}{4a^2}</math>.
-Factoring the LHS gives
-<math>\left(x+\frac{b}{2a}\right)^2=\frac{b^2-4ac}{4a^2}</math>
-As described above, an equation in this form can be solved, yielding
-<math>{x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}}</math>
-This formula is also called the [[Quadratic Formula]].
-We simply plug in a, b, and c and out pops the 2 values of x.

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Difference between revisions of "Quadratic equation"

Revision as of 20:59, 17 June 2006

Contents

Quadratic Equations

Factoring

Completing the square

Quadratic Formula