Solving Equations
1.5 Solving Quadratics
Quadratic Equations and Roots
Quadratic equations are everywhere in algebra, and the main goal is to find all roots.
Definition: Quadratic Equation
A quadratic equation has the form
$$ax^2+bx+c=0,\qquad a\ne0.$$
Theorem: Zero Product Property
If $AB=0$, then $A=0$ or $B=0$ (or both).
Example: Solve by factoring and zero product
Solve:
$$x^2-x-6=0.$$
Show Solution
Factor:
$$x^2-x-6=(x-3)(x+2).$$
Set each factor equal to zero:
$$x-3=0\Rightarrow x=3,$$ $$x+2=0\Rightarrow x=-2.$$
Final solution set:
$${-2,3}. $$
MyOpenMath: Practice: solving quadratics by factoring (placeholder)
Square Root Property
If a quadratic is already in squared form, use roots directly.
Fact: Square Root Property
If
$$x^2=a,$$
then
$$x=\pm\sqrt{a}.$$
Example: Solve with square root property
Solve:
$$2x^2-18=0.$$
Show Solution
Isolate $x^2$:
$$2x^2=18\Rightarrow x^2=9.$$
Apply square root property:
$$x=\pm 3.$$
Final solution set:
$${-3,3}. $$
Example: Solve a shifted square
Solve:
$$(x-4)^2=13.$$
Show Solution
Take square roots of both sides:
$$x-4=\pm\sqrt{13}. $$
Add 4:
$$x=4\pm\sqrt{13}. $$
Final solutions:
$$x=4+\sqrt{13},;x=4-\sqrt{13}. $$
MyOpenMath: Practice: square-root-property quadratics (placeholder)
Quadratic Formula
When factoring is inconvenient or impossible over integers, the quadratic formula always works.
Theorem: Quadratic Formula
For $ax^2+bx+c=0$ with $a\ne 0$,
$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}. $$
Example: Solve with quadratic formula
Solve:
$$2x^2+3x-5=0.$$
Show Solution
Here $a=2$, $b=3$, $c=-5$.
Compute discriminant:
$$b^2-4ac=3^2-4(2)(-5)=9+40=49.$$
Substitute:
$$x=\frac{-3\pm\sqrt{49}}{4}=\frac{-3\pm 7}{4}. $$
So
$$x=1\quad\text{or}\quad x=-\frac{5}{2}. $$
Final solution set:
$$\left{-\frac{5}{2},1\right}. $$
