Topic 6 Quadratic Formula

6.1 Estimate a Square Root

Can you estimate the irrational numbers $\sqrt{2}$ , $\sqrt{3}$ , $\sqrt{5}$ and $\sqrt{7}$ without using a calculator?

Can you estimate the square root $\sqrt{m^{2} + n}$ , where $m$ and $n$ are positive integers?

6.2 Completing the Square

The square root property:
Suppose that $X^{2} = d$ . Then $X = \sqrt{d}$ or $X = - \sqrt{d}$ , or simply $X = \pm \sqrt{d}$ .

The square root property provides another method to solve a quadratic equation, completing the square. This method is based on the following observations: $x^{2} + b x + {(\frac{b}{2})}^{2} = {(x + \frac{b}{2})}^{2},$ and more generally, let $f (x) = a x^{2} + b x + c$ , and $h = - \frac{b}{2 a}$ , then $a x^{2} + b x + c = a (x - h)^{2} + f (h) = a {(x + \frac{b}{2 a})}^{2} + \frac{4 a c - b^{2}}{4 a^{2}} .$

The procedure to rewrite a trinomial as the sum of a perfect square and a constant is called completing the square.

Example 6.1 Solve the equation $x^{2} + 2 x - 1 = 0$ .

Solution.

Isolate the constant. $x^{2} + 2 x = 1$
With $b = 2$ , add ${(\frac{2}{2})}^{2}$ to both sides to complete a square for the binomial $x^{2} + b x$ . $\begin{aligned} x^{2} + 2 x + {(\frac{2}{2})}^{2} & = 1 + {(\frac{2}{2})}^{2} \\ {(x + \frac{2}{2})}^{2} & = 1 + 1 \\ (x + 1)^{2} & = 2 \end{aligned}$
Solve the resulting equation using the square root property. $\begin{aligned} x + 1 = \sqrt{2} & or & x + 1 = - \sqrt{2} \\ x = - 1 + \sqrt{2} & or & x = - 1 - \sqrt{2} \end{aligned}$

Note that the solution can also be written as $x = - 1 \pm \sqrt{2}$ .

Example 6.2 Solve the equation $- 2 x^{2} + 8 x - 9 = 0$ .

Solution.

Isolate the constant. $- 2 x^{2} + 8 x = 9$
Divide by $- 2$ to rewrite the equation in $x^{2} + B x = C$ form $x^{2} - 4 x = - \frac{9}{2}$
With $b = - 4$ , add ${(\frac{- 4}{2})}^{2} = 4$ to both sides to complete the square for the binomial $x^{2} - 4 x$ . $\begin{aligned} x^{2} - 4 x + 4 & = - \frac{9}{2} + 4 \\ (x - 2)^{2} & = - \frac{1}{2} \end{aligned}$
Solve the resulting equation and simplify. $\begin{aligned} x - 2 = \frac{i}{\sqrt{2}} & or & x - 2 = - \frac{i}{\sqrt{2}} \\ x = 2 + \frac{\sqrt{2}}{2} i & or & x = 2 - \frac{\sqrt{2}}{2} i \end{aligned}$

Another way to complete the square is to use the formula $a x^{2} + b x + c = a (x - h)^{2} + f (h)$ , where $f (h) = a h^{2} + b h + c$ is the value of the polynomial $a x^{2} + b x + c$ at $x = h$ .

6.3 The Quadratic Formula

Using the method of completing the square, we obtain the follow quadratic formula for the quadratic equation $a x^{2} + b x + c = 0$ with $a \neq 0$ : $x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} .$

The quantity $b^{2} - 4 a c$ is called the discriminant of the quadratic equation.

If $b^{2} - 4 a c > 0$ , the equation has two real solutions.
If $b^{2} - 4 a c = 0$ , the equation has one real solution.
If $b^{2} - 4 a c < 0$ , the equation has two imaginary solutions (no real solutions).

Example 6.3 Determine the type and the number of solutions of the equation $(x - 1) (x + 2) = - 3$ .

Solution.

Rewrite the equation in the form $a x^{2} + b x + c = 0$ . $\begin{aligned} (x - 1) (x + 2) & = - 3 \\ x^{2} + x + 1 & = 0 \end{aligned}$
Find the values of $a$ , $b$ and $c$ . $a = 1, b = 1 and c = 1.$
Find the discriminant $b^{2} - 4 a c$ . $b^{2} - 4 a c = 1^{2} - 4 \cdot 1 \cdot 1 = - 3.$

The equation has two imaginary solutions.

Example 6.4 Solve the equation $2 x^{2} - 4 x + 7 = 0$ .

Solution.

Find the values of $a$ , $b$ and $c$ . $a = 2, b = - 4 and c = 7.$
Find the discriminant $b^{2} - 4 a c$ . $b^{2} - 4 a c = (- 4)^{2} - 4 \cdot 2 \cdot 7 = 16 - 56 = - 40.$
Apply the quadratic formula and simplify. $\begin{aligned} x = & \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} \\ = & \frac{- (- 4) \pm \sqrt{- 40}}{2 \cdot 2} \\ = & \frac{4 \pm 2 \sqrt{10} i}{4} \\ = & 1 \pm \frac{\sqrt{10}}{2} i . \end{aligned}$

Example 6.5 Find the base and the height of a triangle whose base is three inches more than twice its height and whose area is $5$ square inches. Round your answer to the nearest tenth of an inch.

Solution.

We may suppose the height is $x$ inches. The base can be expressed as $2 x + 3$ inches.
By the area formula for a triangle, we have an equation. $\frac{1}{2} x (2 x + 3) = 5.$
Rewrite the equation in $a x^{2} + b x + c = 0$ form. $\begin{aligned} x (2 x + 3) & = 10 \\ 2 x^{2} + 3 x - 10 & = 0. \end{aligned}$
By the quadratic formula, we have $x = \frac{- 3 \pm \sqrt{3^{2} - 4 \cdot 2 \cdot (- 10)}}{2 \cdot 2} = \frac{- 3 \pm \sqrt{89}}{4} .$

Since $x$ can not be negative, $x = \frac{- 3 + \sqrt{89}}{4} \approx 1.6$ and $2 x + 3 \approx 6.2$ . The height and base of the triangle are approximately $1.6$ inches and 6.2 inches respectively.

6.4 Practice

Problem 6.1 Solve the quadratic equation by the square root property.

$2 x^{2} - 6 = 0$
$(x - 3)^{2} = 10$
$4 (x + 1)^{2} + 25 = 0$

Problem 6.2 Solve the quadratic equation by completing the square.

$x^{2} + x - 1 = 0$
$x^{2} + 8 x + 12 = 0$
$3 x^{2} + 6 x - 1 = 0$

Problem 6.3 Determine the number and the type of solutions of the given equation.

$x^{2} + 8 x + 3 = 0$
$3 x^{2} - 2 x + 4 = 0$
$2 x^{2} - 4 x + 2 = 0$

Problem 6.4 Solve using the quadratic formula.

$x^{2} + 3 x - 7 = 0$
$2 x^{2} = - 4 x + 5$
$2 x^{2} = x - 3$

Problem 6.5 Solve using the quadratic formula.

$(x - 1) (x + 2) = 3$
$2 x^{2} - x = (x + 2) (x - 2)$
$\frac{1}{2} x^{2} + x = \frac{1}{3}$

Problem 6.6 A triangle whose area is $7.5$ square meters has a base that is one meter less than triple the height. Find the length of its base and height. Round to the nearest hundredth of a meter.

Problem 6.7 A rectangular garden whose length is $2$ feet longer than its width has an area 66 square feet. Find the dimensions of the garden, rounded to the nearest hundredth of a foot.