Solving Quadratics — AP® Precalculus Lesson

Part 1 · Hook

Why this matters

Throw a ball and its height traces a parabola: up, a highest point, then down. Quadratic functions describe that shape and many others, from the area of a garden to a company's profit. Topic 1.3 measures how their rates of change behave, so the algebra of quadratics comes first.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. What is $(x - 2)(x - 3)$ multiplied out?

$x^2 - 5x + 6$
$x^2 + 5x + 6$
$x^2 - 6$
$x^2 - 5x - 6$

Show the answer

$x^2 - 3x - 2x + 6$ .

Correct: $x^2 - 5x + 6$ :
$x^2 + 5x + 6$ :
$x^2 - 6$ :
$x^2 - 5x - 6$ :

2. The line through $(0, 7)$ and $(4, 7)$ has slope

$0$
$7$
$\frac{7}{4}$
undefined

Show the answer

The outputs are equal, so the rise is $0$ .

Correct: $0$ :
$7$ :
$\frac{7}{4}$ :
undefined:

Part 4 · See it

See it first

The parabola y = x² − 2x − 3 = (x − 1)² − 4 opens upward. Its vertex is (1, −4), on the dashed axis of symmetry x = 1. Its zeros are x = −1 and x = 3, and its y-intercept is (0, −3). — $y$ $= x^2 - 2x - 3$ $= (x - 1)^2 - 4$ : vertex $(1, -4)$ on the axis of symmetry $x = 1$ , zeros $-1$ and $3$ . LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

A quadratic is written as a product of two factors set to $0$ each factor equal to $0$ gives a solution
The quadratic does not split into easy factorsthe quadratic formula $x$ $= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ gives the solutions
The discriminant $b^2 - 4ac$ is positive, zero or negativethere are two, one or no real zeros
The quadratic is in vertex form $a(x - h)^2 + k$ the vertex is $(h, k)$ , the axis is $x = h$ , and $a > 0$ opens up, $a < 0$ opens down

Part 6 · Key ideas

Key ideas

A quadratic function is $f(x) = ax^2 + bx + c$ with $a \ne 0$ ; its graph is a parabola.
Solve by factors, by square roots ( $(x - 1)^2 = 4$ gives $x - 1 = \pm 2$ ) or by the quadratic formula.
The vertex is at $x = -\frac{b}{2a}$ ; completing the square gives vertex form $a(x - h)^2 + k$ .

Part 7 · Misconception

A common mistake

The wrong idea: $(x - 1)^2 = 4$ means $x - 1 = 2$ , so $x = 3$ .

What actually happens: Both $2^2$ and $(-2)^2$ equal $4$ , so $x - 1 = 2$ or $x - 1 = -2$ : $x = 3$ or $x = -1$ . Taking a square root gives $\pm$ .

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

1. Solve $x^2 - 5x + 6 = 0$ .

$x = -2$ or $x = -3$
$x = 5$ or $x = 6$
$x = 2$ or $x = 3$
$x = 1$ or $x = 6$

Show the answer

Find two numbers with product $6$ and sum $-5$ : $-2$ and $-3$ . So $(x - 2)(x - 3) = 0$ , giving $x = 2$ or $x = 3$ .

$x = -2$ or $x = -3$ : $(x + 2)(x + 3)$ $= x^2 + 5x + 6$ , which has $+5x$ . The signs in the factors are minus.
$x = 5$ or $x = 6$ : These are the coefficients, not solutions: $5^2 - 25 + 6 = 6 \ne 0$ .
Correct: $x = 2$ or $x = 3$ : Right: $x^2 - 5x + 6$ $= (x - 2)(x - 3)$ , and a product is $0$ when a factor is $0$ .
$x = 1$ or $x = 6$ : $1 \cdot 6 = 6$ , but $1 + 6 = 7$ , not $5$ .

2. Use the quadratic formula to find the larger solution of $2x^2 + 3x - 4 = 0$ . Give the exact value.

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

$a = 2$ , $b = 3$ , $c = -4$ . $x$ $= \frac{-3 \pm \sqrt{9 - 4(2)(-4)}}{2(2)}$ $= \frac{-3 \pm \sqrt{41}}{4}$ . The larger is $\frac{-3 + \sqrt{41}}{4} \approx 0.851$ .

Answer: $\frac{-3 + \sqrt{41}}{4}$ ≈ 0.851

3. What is the vertex of the parabola $y = 2x^2 - 8x + 5$ ?

$(2, -3)$
$(-2, 29)$
$(4, 5)$
$(2, 5)$

Show the answer

The vertex is at $x = -\frac{b}{2a} = -\frac{-8}{4} = 2$ , and $y$ $= 2(2)^2 - 8(2) + 5$ $= -3$ .

Correct: $(2, -3)$ : Right: $x = -\frac{-8}{2 \cdot 2} = 2$ and $y = 8 - 16 + 5 = -3$ .
$(-2, 29)$ : The vertex is at $x = -\frac{b}{2a} = -\frac{-8}{4} = 2$ , not $-2$ .
$(4, 5)$ : This uses $x = -\frac{b}{a}$ ; divide by $2a$ .
$(2, 5)$ : $5$ is the $y$ -intercept; put $x = 2$ into the formula.

4. How many real zeros does $f(x) = 3x^2 - 2x + 5$ have?

Two
One
Zero
It depends on $x$

Show the answer

The discriminant $b^2 - 4ac = 4 - 60 = -56$ is negative, so the square root in the quadratic formula is not real: no real zeros. The parabola opens up and stays above the $x$ -axis.

Two: Two real zeros need $b^2 - 4ac > 0$ ; here it is $-56$ .
One: One real zero needs $b^2 - 4ac = 0$ .
Correct: Zero: Right: the discriminant is $(-2)^2 - 4(3)(5)$ $= 4 - 60$ $= -56 < 0$ .
It depends on $x$ : The number of zeros is a property of the whole function, found from $a$ , $b$ and $c$ .

5. A parabola has vertex $(-1, 2)$ and passes through $(0, 5)$ . Write its equation in vertex form.

Type an expression for $y =$ .

Show the answer

Vertex form $y = a(x - h)^2 + k$ with $(h, k) = (-1, 2)$ : $y = a(x + 1)^2 + 2$ . The point $(0, 5)$ gives $5 = a + 2$ , so $a = 3$ .

Answer: $y =$ $3(x + 1)^{2} + 2$

6. What are the zeros of $f(x) = (x - 1)^2 - 4$ ?

$x = 1$ and $x = -4$
$x = -1$ and $x = 3$
$x = 3$
$x = 5$ and $x = -3$

Show the answer

Set $(x - 1)^2 - 4 = 0$ : $(x - 1)^2 = 4$ , $x - 1 = 2$ or $x - 1 = -2$ , so $x = 3$ or $x = -1$ .

$x = 1$ and $x = -4$ : These come from the vertex $(1, -4)$ , not from setting $f(x) = 0$ .
Correct: $x = -1$ and $x = 3$ : Right: $(x - 1)^2 = 4$ , so $x - 1 = \pm 2$ .
$x = 3$ : $(x - 1)^2 = 4$ has two square roots, $2$ and $-2$ : $x = 3$ or $x = -1$ .
$x = 5$ and $x = -3$ : These come from $x - 1 = \pm 4$ , which skips the square root. Take the square root of $4$ : $x - 1 = \pm 2$ .

Part 9 · Summary

Summary

A quadratic

f(x) = ax^2 + bx + c

graphs as a parabola with its vertex at

x = -\frac{b}{2a}

. Solve

f(x) = 0

by factors, by square roots or by the quadratic formula; the discriminant

b^2 - 4ac

counts the real zeros. Completing the square gives vertex form

a(x - h)^2 + k

, which shows the vertex

(h, k)

.

Part 10 · Up next

What comes next

Part 11 · Connections