3-03 Solve Quadratic Equations by Graphing and Finding Square Roots (3.1)

Example 1: Solve by Graphing

Solve 0 = 2x² + 4x − 6 by graphing.

Solution

The equation already equals zero.

Graph the function. Since this is general form, the x-value of the vertex is \(x=-\frac{b}{2a}\).

$$ x=\frac{-4}{2\left(2\right)}=-1 $$

Make a table of values.

x	−4	−3	−2	−1	0	1	2
y	10	0	−6	−8	−6	0	10

Plot the points and graph the function.

The solutions to the equations are the x-values of the x-intercepts.

x = −3 or 1

Example 2: Solve by Graphing

Solve 2 = −x² − 5x − 2 by graphing.

Solution

Make the equation equal zero.

2 = −x² − 5x − 2

0 = −x² − 5x − 4

Graph the function. Since this is general form, the x-value of the vertex is \(x=-\frac{b}{2a}\).

$$ x=\frac{5}{2\left(-1\right)}=-\frac{5}{2} $$

Make a table of values.

x	−7	−6	−5	−4	−3	−5/2	−2	−1	0	1	2
y	−18	−10	−4	0	2	2.25	2	0	−4	−10	−18

Plot the points and graph the function.

The solutions to the equations are the x-values of the x-intercepts.

x = 4 or −1

Example 3: Solve by Square Roots

Solve 3x² − 5 = 7.

Solution

The only time x appears in the equation is as x², so solve by finding square roots.

Solve for the squared expression, in this case x².

3x² − 5 = 7

3x² = 12

x² = 4

Take a square root. Remember to put ±.

$$ \sqrt{x^2}=\pm\sqrt4 $$

$$ x=\pm\sqrt4 $$

x = ±2

Example 4: Solve by Square Roots

Solve (x − 1)² + 4 = 16.

Solution

The only time x appears in the equation is (x − 1)², so solve by finding square roots.

Solve for the squared expression, in this case (x − 1)².

(x − 1)² + 4 = 16

(x − 1)² = 12

Take a square root. Remember to put ±.

$$ \sqrt{\left(x-1\right)^2}=\pm\sqrt{12} $$

x − 1 = ±4·3

$$ x-1=\pm2\sqrt{3} $$

Finish solving for x.

$$ x = 1 ± 2 \sqrt{3} $$

Example 5: Solve by Square Roots

Solve 2(x + 2)² − 1 = −19.

Solution

The only time x appears in the equation is (x + 2)², so solve by finding square roots.

Solve for the squared expression, in this case (x + 2)².

2(x + 2)² − 1 = −19

2(x + 2)² = −18

(x + 2)² = −9

Take a square root. Remember to put ±.

$$ \sqrt{\left(x+2\right)^2}=\pm\sqrt{-9} $$

x + 2 = ±3i

Finish solving for x.

x = −2 ± 3i

Example 6: Maximize Revenue

A video game store charges $30 per game and sells 40 games each day. They try dropping the price by $1 and sell 2 more games a day. How much should the video game store charge to maximize their daily revenue? What is their maximum daily revenue?

Solution

Revenue is price × number sold.

R = p × s

In this case the price is (30 − x) where x is the number of times $1 is deducted from the price. The number sold is (40 + 2x). Write the formula for revenue.

R = (30 − x)(40 + 2x)

Use the distributive property to put this in general form.

R = 30·40 + 30·2x − x·40 − x·2x

R = 120 + 60x − 40x − 2x²

R = −2x² + 20x + 120

The maximum is at the vertex. In general form, the x-value of the vertex is \(x=-\frac{b}{2a}\).

$$ x=-\frac{20}{2\left(-2\right)}=5 $$

Remember x is the number of times to lower the price by $1. So, the store should lower the price by 5×$1 = $5. They should charge (30 – x) = (30 – 5) = $25.

The maximum revenue is the y-value of the vertex. Substitute x = 5 into the revenue function.

R = –2(5)² + 20(5) + 120 = $170

The store should charge $25 per game to make a revenue of $170 per day.