Lesson 5: Chapter Summary and Review

New Concept

Quadratic equations involve a squared variable, like $x^2$. This chapter review connects the standard form $ax^2+bx+c=0$ to its U-shaped graph (a parabola) and the key methods used to find its solutions.

What’s next

Next, you'll use our interactive examples and practice cards to master solving these equations and graphing their corresponding parabolas.

Property

A quadratic equation has the standard form $ax^2 + bx + c = 0$, where $a$, $b$, and $c$ are constants and $a$ is not equal to zero.

Examples

• To write $3x^2 = 5x - 2$ in standard form, move all terms to one side to get $3x^2 - 5x + 2 = 0$. Here, $a=3$, $b=-5$, and $c=2$.

• The equation $x(x+4) = 5$ is quadratic. First, distribute to get $x^2 + 4x = 5$, then rearrange into standard form: $x^2 + 4x - 5 = 0$.

Property

To solve a quadratic equation of the form $ax^2 + c = 0$:

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

Every quadratic equation has two solutions, which may be the same.

Examples

• To solve $2x^2 - 50 = 0$, first isolate $x^2$ by adding 50 and dividing by 2 to get $x^2 = 25$. Then, take the square root of both sides to find the solutions $x = 5$ and $x = -5$.

Property

The product of two factors equals zero if and only if one or both of the factors equals zero. In symbols, $ab = 0$ if and only if $a = 0$ or $b = 0$ or both.

To Solve a Quadratic Equation by Factoring:

1. Write the equation in standard form.
2. Factor the left side of the equation.
3. Apply the zero-factor principle: Set each factor equal to zero.
4. Solve each equation.

Examples

• To solve $x^2 - 3x - 10 = 0$, factor the left side into $(x-5)(x+2) = 0$. By the zero-factor principle, either $x-5=0$ or $x+2=0$. The solutions are $x=5$ and $x=-2$.

Property

The graph of a quadratic equation $y = ax^2 + bx + c$ is called a parabola.
For the graph of $y = ax^2$: The parabola opens upward if $a > 0$ and downward if $a < 0$. The magnitude of $a$ determines how wide or narrow the parabola is.
For the graph of $y = x^2 + c$: The graph is shifted upward by $c$ units if $c > 0$ and downward if $c < 0$.
For the graph of $y = ax^2 + bx + c$, the $x$-coordinate of the vertex is $x_v = \frac{-b}{2a}$.

Examples

• For the parabola $y = x^2 - 6x + 5$, the vertex's x-coordinate is $x_v = \frac{-(-6)}{2(1)} = 3$. To find the y-coordinate, plug $x=3$ back in: $y = (3)^2 - 6(3) + 5 = 9 - 18 + 5 = -4$. The vertex is at $(3, -4)$.

• The graph of $y = -3x^2$ opens downward because $a=-3$ is less than 0. It is also narrower than the basic parabola $y=x^2$ because the magnitude of $a$ is greater than 1.

Property

To Solve a Quadratic Equation by Completing the Square:

1. Write the equation in standard form. Divide both sides by the coefficient of the quadratic term, and subtract the constant term from both sides.
2. Complete the square on the left side: Multiply the coefficient of the first-degree term by one-half, then square the result. Add this value to both sides.
3. Write the left side of the equation as the square of a binomial. Simplify the right side.
4. Use extraction of roots to finish the solution.

Examples

• To solve $x^2 + 6x - 7 = 0$, move the 7 over: $x^2 + 6x = 7$. Half of 6 is 3, and $3^2=9$. Add 9 to both sides: $x^2 + 6x + 9 = 16$. Factor the left side: $(x+3)^2 = 16$. So, $x+3 = \pm 4$, which gives $x=1$ and $x=-7$.

• Solve $x^2 - 10x + 20 = 0$. Move 20 to get $x^2 - 10x = -20$. Half of -10 is -5, and $(-5)^2=25$. Add 25 to both sides: $x^2 - 10x + 25 = 5$. This becomes $(x-5)^2 = 5$, so the exact solutions are $x = 5 \pm \sqrt{5}$.