Lesson 6: Chapter Summary and Review

New Concept

A quadratic equation, in the standard form $ax^2 + bx + c = 0$, is a core algebraic concept. This lesson introduces various methods to solve these equations and analyze their corresponding graphs, which are called parabolas.

What’s next

You're about to dive into worked examples, interactive graphs, and practice problems to master solving and graphing these equations. Let's get started!

Property

A quadratic equation can be written in the standard form

1. Isolate $x^2$ on one side of the equation.
2. Take the square root of each side.

Examples

• To solve $2x^2 - 50 = 0$, first add 50 to both sides to get $2x^2 = 50$. Then divide by 2 to get $x^2 = 25$. Taking the square root of both sides gives the solutions $x=5$ and $x=-5$.

• To solve $3y^2 - 12 = 0$, isolate the $y^2$ term by adding 12 to both sides, which gives $3y^2 = 12$. Divide by 3 to get $y^2 = 4$. The solutions are $y=2$ and $y=-2$.

Property

Zero-Factor Principle: If the product of two numbers is zero, then one (or both) of the numbers must be zero. If $AB = 0$, then either $A = 0$ or $B = 0$.

To Solve a Quadratic Equation by Factoring:

1. Write the equation in standard form, $ax^2 + bx + c = 0$.
2. Factor the left side of the equation.
3. Apply the Zero-Factor Principle; set each factor equal to zero.
4. Solve each equation to obtain two solutions.

Examples

• To solve $x^2 + 2x - 15 = 0$, find two numbers that multiply to $-15$ and add to $2$ (which are $5$ and $-3$). Factoring gives $(x+5)(x-3)=0$. The solutions are $x=-5$ and $x=3$.

Property

To Graph a Quadratic Equation:

1. Find the $x$-intercepts: set $y = 0$ and solve for $x$.
2. Find the vertex: the $x$-coordinate is the average of the $x$-intercepts. Find the $y$-coordinate by substituting the $x$-coordinate into the equation of the parabola.
3. Find the $y$-intercept: set $x = 0$ and solve for $y$.
4. Draw a parabola through the points.

Examples

• To find the vertex of $y = x^2 - 8x + 12$, first find the x-intercepts by solving $x^2 - 8x + 12 = 0$, which gives $(x-2)(x-6)=0$, so $x=2$ and $x=6$. The vertex's x-coordinate is $\frac{2+6}{2} = 4$. The y-coordinate is $4^2 - 8(4) + 12 = -4$. The vertex is $(4, -4)$.

• The $x$-intercepts of $y = x^2 - 25$ are found where $y=0$, so $x^2 = 25$, which means $x = 5$ and $x = -5$. The vertex's x-coordinate is $\frac{5+(-5)}{2} = 0$. The y-coordinate is $0^2-25 = -25$. The vertex is $(0, -25)$.

Property

The solutions of the equation

Examples

• To solve $x^2 + 4x - 5 = 0$, use $a=1, b=4, c=-5$. The formula gives $x = \frac{-4 \pm \sqrt{4^2 - 4(1)(-5)}}{2(1)} = \frac{-4 \pm \sqrt{16+20}}{2} = \frac{-4 \pm 6}{2}$. The solutions are $x=1$ and $x=-5$.

• To solve $3x^2 - 2x - 2 = 0$, use $a=3, b=-2, c=-2$. The formula gives $x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(3)(-2)}}{2(3)} = \frac{2 \pm \sqrt{4+24}}{6} = \frac{2 \pm \sqrt{28}}{6}$.