Lesson 5: Chapter Summary and Review

New Concept

The quadratic formula is a universal tool for solving equations of the form $ax^2 + bx + c = 0$. We'll use it to find a parabola's x-intercepts and understand the nature of its solutions.

What’s next

Now, you'll apply this formula by working through interactive examples, a series of practice cards, and challenge problems to solidify your skills.

Property

The solutions of the equation $ax^2 + bx + c = 0$, $(a \neq 0)$ are

Examples

• To solve $x^2 - 3x + 1 = 0$, use $a=1, b=-3, c=1$. The solutions are $x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(1)}}{2(1)} = \frac{3 \pm \sqrt{5}}{2}$.

• To solve $4x^2 + 12x + 9 = 0$, use $a=4, b=12, c=9$. The solution is $x = \frac{-12 \pm \sqrt{12^2 - 4(4)(9)}}{2(4)} = \frac{-12 \pm \sqrt{0}}{8} = -\frac{3}{2}$.

Property

The discriminant of a quadratic equation is

1. If $D > 0$, there are two unequal real solutions.
2. If $D = 0$, there is one solution of multiplicity two.
3. If $D < 0$, there are two complex conjugate solutions.

Examples

• For $2y^2 - 3y - 4 = 0$, the discriminant is $D = (-3)^2 - 4(2)(-4) = 9 + 32 = 41$. Since $D > 0$, there are two distinct real solutions.

• For $4x^2 - 12x + 9 = 0$, the discriminant is $D = (-12)^2 - 4(4)(9) = 144 - 144 = 0$. Since $D = 0$, there is one repeated real solution.

Property

A quadratic equation $y = ax^2 + bx + c$, $a \neq 0$, can be written in the vertex form

where the vertex of the graph is $(x_v, y_v)$.

Examples

• The equation $y = -2(x - 1)^2 + 5$ is in vertex form. The vertex is at $(1, 5)$, and because $a = -2$ is negative, the parabola opens downward.

• To write $y = x^2 - 8x + 10$ in vertex form, find the vertex. $x_v = \frac{-(-8)}{2(1)} = 4$. Then $y_v = 4^2 - 8(4) + 10 = -6$. The vertex form is $y = (x - 4)^2 - 6$.

Property

1. Write the inequality in standard form: One side is zero, and the other has the form $ax^2 + bx + c$.
2. Find the $x$-intercepts of the graph of $y = ax^2 + bx + c$ by setting $y = 0$ and solving for $x$.
3. Make a rough sketch of the graph, using the sign of $a$ to determine whether the parabola opens upward or downward.
4. Decide which intervals on the $x$-axis give the correct sign for $y$.

Examples

• To solve $(x - 3)(x + 2) > 0$, the intercepts are $x=3$ and $x=-2$. The parabola opens up, so it's positive when $x < -2$ or $x > 3$. The solution is $(-\infty, -2) \cup (3, \infty)$.

• To solve $y^2 - y - 12 \leq 0$, factor to get $(y-4)(y+3) \leq 0$. The intercepts are $y=4$ and $y=-3$. The parabola opens up, so it is negative or zero between the intercepts. The solution is $[-3, 4]$.

Property

1. The closed interval $[a, b]$ is the set $a \leq x \leq b$.
2. The open interval $(a, b)$ is the set $a < x < b$.
3. Intervals may also be half-open or half-closed.
4. The infinite interval $[a, \infty)$ is the set $x \geq a$.
5. The infinite interval $(-\infty, a]$ is the set $x \leq a$.

Examples

• The inequality $x > -5$ represents all numbers greater than -5. In interval notation, this is written as the open infinite interval $(-5, \infty)$.

• The compound inequality $-1 \leq x < 8$ describes a half-open interval that includes $-1$ but excludes $8$. This is written as $[-1, 8)$.