Algebra/Quadratic Equation

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Algebra
Algebra
 ← Completing the Square Quadratic Equation Binomial Theorem → 

Derivation[edit]

The solutions to the general-form quadratic function ax^2 + bx + c = 0 can be given by a simple equation called the quadratic equation. To solve this equation, recall the completed square form of the quadratic equation derived in the previous section:

 y = a(x+\frac{b}{2a})^2 + c - \frac{b^2}{4a}

In this case,  y = 0 since we're looking for the root of this function. To solve, first subtract c and divide by a:

 -\frac{c}{a} + \frac{b^2}{4a^2} = (x + \frac{b}{2a})^2

Take the (plus and minus) square root of both sides to obtain:

 \pm \sqrt{-\frac{c}{a} + \frac{b^2}{4a^2}} = x + \frac{b}{2a}

Subtracting  \frac{b}{2a} from both sides:

 x = \frac{-b}{2a} \pm \sqrt{-\frac{c}{a} + \frac{b^2}{4a^2}}

This is the solution but it's in an inconvenient form. Let's rationalize the denominator of the square root:

 \sqrt{-\frac{c}{a} + \frac{b^2}{4a^2}} =  \sqrt{\frac{-4ac + b^2}{4a^2}} = \frac{\sqrt{b^2-4ac}}{2|a|} = \pm \frac{\sqrt{b^2-4ac}}{2a}

Now, adding the fractions, the final version of the quadratic formula is:

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

This formula is very useful, and it is suggested that the students memorize it as soon as they can.

Discriminant[edit]

The part under the radical sign, {b^2-4ac}, is called the discriminant, Δ. The value of the discriminant tells us some useful information about the roots.

  • If Δ > 0, there are two unique real solutions.
  • If Δ = 0, there is one unique real solution.
  • If Δ < 0, there are two unique, conjugate imaginary solutions.
  • If Δ is a perfect square then the two solutions are rational, otherwise they are irrational conjugates.

Word Problems[edit]

Need to pull word problems from http://teachers.yale.edu/curriculum/search/viewer.php?id=initiative_07.06.12_u&skin=h