Algebra II/Quadratic Functions

A quadratic function is represented by the following equation:
 * $$y = ax$$2$$ + bx + c$$


 * $$ax$$2 = Quadratic Term
 * $$bx$$ = Linear Term
 * $$c$$ = Constant Term

Solving Quadratic Functions by Factoring
1. $$x$$2 $$+ 7x + 6$$


 * Factor them: We get $$(x + 6)(x + 1)$$.
 * The linear terms must add to make 7.
 * The constant terms needs to multiply to make 6.
 * Set them out as problems to solve:
 * $$x + 6 = 0$$ → x = -6
 * $$x + 1 = 0$$ → x = -1
 * Your answers are $$-6$$ and $$-1$$.

2. $$x$$2 $$- 9$$.


 * Factor them: We get $$(x - 3)(x + 3)$$.
 * Set them out as problems to solve:
 * $$x + 3 = 0$$ → x = -3
 * $$x - 3 = 0$$ → x = 3
 * Your answers are $$3$$ and $$-3$$.

3. $$2x$$2$$ + 12x + 10$$


 * Divide all of the terms by the GCF: 2: We get a new problem to deal with, which is $$x$$2$$ + 6x + 5$$.
 * Factor them: We get $$2(x + 5)(x + 1)$$.
 * Set them out as problems to solve:
 * $$x + 5 = 0$$ → x = -5
 * $$x + 1 = 0$$ → x = -1
 * Your answers are $$-5$$ and $$-1$$.

Solving Quadratic Factors by Completing the Square
1. $$x$$2 $$+ 8x$$ __ = $$19$$


 * Take the Linear Term and divide it by $$2$$: We get $$4$$.
 * We take this number, $$4$$, and square it: We get $$16$$.
 * We add $$16$$ to $$19$$: We get $$35$$.
 * We now have: $$(x + 4)$$2 $$= 35$$.
 * We square both sides: We get $$x + 4 = $$√$$35$$.
 * We minus 4 to the other side. Here is our answer.: $$x = -4 $$± √$$35$$.

2. $$x$$2 $$+ x$$ + ___


 * Take the Linear Term and divide it by $$2$$: We get $$\tfrac{1}{2}$$.
 * We take this number, $$\tfrac{1}{2}$$, and square it: We get $$\tfrac{1}{4}$$.
 * We have our answer: $$x$$2 $$+ x$$ + $$\tfrac{1}{4}$$.

3. $$x$$2 $$+ 45$$ = $$10x$$


 * Rearrange this problem so that it matches the standard format for a quadratic equation: We switch the $$45$$ and the $$10x$$ around, forming our new problem: $$x$$2 $$- 10x$$ = $$-45$$.
 * Divide the Linear Tearm, $$bx$$, by $$2$$: This gives us $$5$$.
 * Square the $$5$$: This gives us $$25$$.
 * Add the $$25$$ to $$-45$$: This brings our problem to ($$x$$undefined $$- 5)$$2 = $$-20$$.
 * Square both sides of the problem: This brings us to $$x - 5 = $$i√$$20$$.
 * Find the square root of $$20$$ (don't forget the $$i$$) and then add $$5$$ to the opposite side to find your answer: Our final answer is $$x = -5 $$± $$2i$$√$$5$$.