Solving Equations Quadratic in Form

6.3 Solving Equations Quadratic in Form

Learning Objectives

Develop a general strategy for solving quadratic equations.
Solve equations that are quadratic in form.

General Guidelines for Solving Quadratic Equations

Use the coefficients of a quadratic equation to help decide which method is most appropriate for solving it. While the quadratic formula always works, it is sometimes not the most efficient method. If given any quadratic equation in standard form,

$a x^{2} + b x + c = 0$ where c = 0, then it is best to factor out the GCF and solve by factoring.

Example 1

Solve: $12 x^{2} - 3 x = 0 .$

Solution:

In this case, c = 0 and we can solve by factoring out the GCF 3x.

$\begin{array}{l} 12 x^{2} - 3 x & = & 0 \\ 3 x (4 x - 1) & = & 0 \end{array}$

Then apply the zero-product property and set each factor equal to zero.

$\begin{array}{l} 3 x & = & 0 & or & 4 x - 1 & = & 0 \\ x & = & 0 & 4 x & = & 1 \\ x & = & \frac{1}{4} \end{array}$

Answer: The solutions are 0 and $\frac{1}{4} .$

If b = 0, then we can solve by extracting the roots.

Example 2

Solve: $5 x^{2} + 8 = 0 .$

Solution:

In this case, b = 0 and we can solve by extracting the roots. Begin by isolating the square.

$\begin{array}{l} 5 x^{2} + 8 & = & 0 \\ 5 x^{2} & = & - 8 \\ x^{2} & = & - \frac{8}{5} \end{array}$

Next, apply the square root property. Remember to include the ±.

$\begin{array}{l} x & = & \pm \sqrt{- \frac{8}{5}} & R a t i o n a l i z e t h e d e n o m i n a t o r . \\ = & \pm \frac{\sqrt{- 4 \cdot 2}}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} & S i m p l i f y . \\ = & \pm \frac{2 i \sqrt{2} \cdot \sqrt{5}}{5} \\ = & \pm \frac{2 i \sqrt{10}}{5} \end{array}$

Answer: The solutions are $\pm \frac{2 i \sqrt{10}}{5} .$

When given a quadratic equation in standard form where a, b, and c are all nonzero, determine the value for the discriminant using the formula $b^{2} - 4 a c .$

If the discriminant is a perfect square, then solve by factoring.
If the discriminant is not a perfect square, then solve using the quadratic formula.

Recall that if the discriminant is not a perfect square and positive, the quadratic equation will have two irrational solutions. And if the discriminant is negative, the quadratic equation will have two complex conjugate solutions.

Example 3

Solve: $(3 x + 5) (3 x + 7) = 6 x + 10 .$

Solution:

Begin by rewriting the quadratic equation in standard form.

$\begin{array}{l} (3 x + 5) (3 x + 7) & = & 6 x + 10 \\ 9 x^{2} + 21 x + 15 x + 35 & = & 6 x + 10 \\ 9 x^{2} + 36 x + 35 & = & 6 x + 10 \\ 9 x^{2} + 30 x + 25 & = & 0 \end{array}$

Substitute a = 9, b = 30, and c = 25 into the discriminant formula.

$\begin{array}{l} b^{2} - 4 a c & = & {(30)}^{2} - 4 (9) (25) \\ = & 900 - 900 \\ = & 0 \end{array}$

Since the discriminant is 0, solve by factoring and expect one real solution, a double root.

$\begin{array}{l} \begin{array}{l} 9 x^{2} + 30 x + 25 & = & 0 \\ (3 x + 5) (3 x + 5) & = & 0 \end{array} \\ \begin{array}{l} 3 x + 5 & = & 0 & or & 3 x + 5 & = & 0 \\ 3 x & = & - 5 & 3 x & = & - 5 \\ x & = & - \frac{5}{3} & x & = & - \frac{5}{3} \end{array} \end{array}$

Answer: The solution is $- \frac{5}{3} .$

It is good to know that the quadratic formula will work to find the solutions to all of the examples in this section. However, it is not always the best solution. If the equation can be solved by factoring or by extracting the roots, you should use that method.

Solving Equations Quadratic in Form

In this section we outline an algebraic technique that is used extensively in mathematics to transform equations into familiar forms. We begin by defining quadratic formAn equation of the form $a u^{2} + b u + c = 0$ where a, b and c are real numbers and u represents an algebraic expression.,

$a u^{2} + b u + c = 0$

Here u represents an algebraic expression. Some examples follow:

$\begin{array}{l} {(\frac{t + 2}{t})}^{2} + 8 (\frac{t + 2}{t}) + 7 & = & 0 & \overset{u = \frac{t + 2}{t}}{\Rightarrow} & u^{2} + 8 u + 7 & = & 0 \\ x^{2 / 3} - 3 x^{1 / 3} - 10 & = & 0 & \overset{u = x^{1 / 3}}{\Rightarrow} & u^{2} - 3 u - 10 & = & 0 \\ 3 y^{- 2} + 7 y^{- 1} - 6 & = & 0 & \overset{u = y^{- 1}}{\Rightarrow} & 3 u^{2} + 7 u - 6 & = & 0 \end{array}$

If we can express an equation in quadratic form, then we can use any of the techniques used to solve quadratic equations. For example, consider the following fourth-degree polynomial equation,

$x^{4} - 4 x^{2} - 32 = 0$

If we let $u = x^{2}$ then $u^{2} = {(x^{2})}^{2} = x^{4}$ and we can write

$\begin{array}{l} x^{4} - 4 x^{2} - 32 & = & 0 & \Rightarrow & {(x^{2})}^{2} & - & 4 (x^{2}) & - & 32 & = & 0 \\ ↓ & ↓ \\ u^{2} & - & 4 u & - & 32 & = & 0 \end{array}$

This substitution transforms the equation into a familiar quadratic equation in terms of u which, in this case, can be solved by factoring.

$\begin{matrix} u^{2} - 4 u - 32 = 0 \\ (u - 8) (u + 4) = 0 \\ u = 8 o r u = - 4 \end{matrix}$

Since $u = x^{2}$ we can back substitute and then solve for x.

$\begin{array}{l} u & = & 8 & or & u & = & - 4 \\ ↓ & ↓ \\ x^{2} & = & 8 & x^{2} & = & - 4 \\ x & = & \pm \sqrt{8} & x & = & \pm \sqrt{- 4} \\ x & = & \pm 2 \sqrt{2} & x & = & \pm 2 i \end{array}$

Therefore, the equation $x^{4} - 4 x^{2} - 32 = 0$ has four solutions ${\pm 2 \sqrt{2}, \pm 2 i}$ , two real and two complex. This technique, often called a u-substitutionA technique in algebra using substitution to transform equations into familiar forms., can also be used to solve some non-polynomial equations.

Example 4

Solve: $x - 2 \sqrt{x} - 8 = 0 .$

Solution:

This is a radical equation that can be written in quadratic form. If we let $u = \sqrt{x}$ then $u^{2} = {(\sqrt{x})}^{2} = x$ and we can write

$\begin{array}{l} x - 2 \sqrt{x} - 8 = 0 \\ ↓ ↓ \\ u^{2} - 2 u - 8 = 0 \end{array}$

Solve for u.

$\begin{matrix} u^{2} - 2 u - 8 = 0 \\ (u - 4) (u + 2) = 0 \\ u = 4 o r u = - 2 \end{matrix}$

Back substitute $u = \sqrt{x}$ and solve for x.

$\begin{array}{l} \sqrt{x} & = & 4 & or & \sqrt{x} & = & - 2 \\ {(\sqrt{x})}^{2} & = & {(4)}^{2} & {(\sqrt{x})}^{2} & = & {(- 2)}^{2} \\ x & = & 16 & x & = & 4 \end{array}$

Recall that squaring both sides of an equation introduces the possibility of extraneous solutions. Therefore we must check our potential solutions.

$\begin{array}{l} C h e c k x = 16 & C h e c k x = 4 \\ \begin{array}{c} x - 2 \sqrt{x} - 8 & = & 0 \\ 16 - 2 \sqrt{16} - 8 & = & 0 \\ 16 - 2 \cdot 4 - 8 & = & 0 \\ 16 - 8 - 8 & = & 0 \\ 0 & = & 0 & ✓ \end{array} & \begin{array}{l} x - 2 \sqrt{x} - 8 & = & 0 \\ 4 - 2 \sqrt{4} - 8 & = & 0 \\ 4 - 2 \cdot 2 - 8 & = & 0 \\ 4 - 4 - 8 & = & 0 \\ - 8 & = & 0 & ✗ \end{array} \end{array}$

Because $x = 4$ is extraneous, there is only one solution, $x = 16 .$

Answer: The solution is 16.

Example 5

Solve: $x^{2 / 3} - 3 x^{1 / 3} - 10 = 0 .$

Solution:

If we let $u = x^{1 / 3}$ , then $u^{2} = {(x^{1 / 3})}^{2} = x^{2 / 3}$ and we can write

$\begin{array}{l} x^{2 / 3} - 3 x^{1 / 3} - 10 = 0 \\ ↓ ↓ \\ u^{2} - 3 u - 10 = 0 \end{array}$

Solve for u.

$\begin{matrix} u^{2} - 3 u - 10 = 0 \\ (u - 5) (u + 2) = 0 \\ u = 5 o r u = - 2 \end{matrix}$

Back substitute $u = x^{1 / 3}$ and solve for x.

$\begin{array}{l} x^{1 / 3} & = & 5 & or & x^{1 / 3} & = & - 2 \\ {(x^{1 / 3})}^{3} & = & {(5)}^{3} & {(x^{1 / 3})}^{3} & = & {(- 2)}^{3} \\ x & = & 125 & x & = & - 8 \end{array}$

Check.

$\begin{array}{l} C h e c k x = 125 & C h e c k x = - 8 \\ \begin{array}{c} x^{2 / 3} - 3 x^{1 / 3} - 10 & = & 0 \\ {(125)}^{2 / 3} - 3 {(125)}^{1 / 3} - 10 & = & 0 \\ {(5^{3})}^{2 / 3} - 3 {(5^{3})}^{1 / 3} - 10 & = & 0 \\ 5^{2} - 3 \cdot 5 - 10 & = & 0 \\ 25 - 15 - 10 & = & 0 \\ 0 & = & 0 & ✓ \end{array} & \begin{array}{c} x^{2 / 3} - 3 x^{1 / 3} - 10 & = & 0 \\ {(- 8)}^{2 / 3} - 3 {(- 8)}^{1 / 3} - 10 & = & 0 \\ {[{(- 2)}^{3}]}^{2 / 3} - 3 {[{(- 2)}^{3}]}^{1 / 3} - 10 & = & 0 \\ {(- 2)}^{2} - 3 \cdot (- 2) - 10 & = & 0 \\ 4 + 6 - 10 & = & 0 \\ 0 & = & 0 & ✓ \end{array} \end{array}$

Answer: The solutions are −8, 125.

Example 6

Solve: $3 y^{- 2} + 7 y^{- 1} - 6 = 0 .$

Solution:

If we let $u = y^{- 1}$ , then $u^{2} = {(y^{- 1})}^{2} = y^{- 2}$ and we can write

$\begin{array}{l} 3 y^{- 2} + 7 y^{- 1} - 6 = 0 \\ ↓ ↓ \\ 3 u^{2} + 7 u - 6 = 0 \end{array}$

Solve for u.

$\begin{matrix} 3 u^{2} + 7 u - 6 = 0 \\ (3 u - 2) (u + 3) = 0 \\ u = \frac{2}{3} o r u = - 3 \end{matrix}$

Back substitute $u = y^{- 1}$ and solve for y.

$\begin{array}{l} y^{- 1} & = & \frac{2}{3} & or & y^{- 1} & = & - 3 \\ \frac{1}{y} & = & \frac{2}{3} & \frac{1}{y} & = & - 3 \\ y & = & \frac{3}{2} & y & = & - \frac{1}{3} \end{array}$

The original equation is actually a rational equation where $y \neq 0 .$ In this case, the solutions are not restrictions; they solve the original equation.

Answer: The solutions are $- \frac{1}{3}, \frac{3}{2} .$

Example 7

Solve: ${(\frac{t + 2}{t})}^{2} + 8 (\frac{t + 2}{t}) + 7 = 0 .$

Solution:

If we let $u = \frac{t + 2}{t}$ , then $u^{2} = {(\frac{t + 2}{t})}^{2}$ and we can write

$\begin{array}{l} {(\frac{t + 2}{t})}^{2} & + & 8 (\frac{t + 2}{t}) & + & 7 & = & 0 \\ ↓ & ↓ \\ u^{2} & + & 8 u & + & 7 & = & 0 \end{array}$

Solve for u.

$\begin{matrix} u^{2} + 8 u + 7 = 0 \\ (u + 1) (u + 7) = 0 \\ u = - 1 o r u = - 7 \end{matrix}$

Back substitute $u = \frac{t + 2}{t}$ and solve for t.

$\begin{array}{l} \frac{t + 2}{t} & = & - 1 & or & \frac{t + 2}{t} & = & - 7 \\ t + 2 & = & - t & t + 2 & = & - 7 t \\ 2 t & = & - 2 & 8 t & = & - 2 \\ t & = & - 1 & t & = & - \frac{1}{4} \end{array}$

Answer: The solutions are −1, $- \frac{1}{4} .$ The check is left to the reader.

Try this! Solve: $12 x^{- 2} - 16 x^{- 1} + 5 = 0$

Answer: The solutions are $\frac{6}{5}$ , 2.

(click to see video)

So far all of the examples were of equations that factor. As we know, not all quadratic equations factor. If this is the case, we use the quadratic formula.

Example 8

Solve: $x^{4} - 10 x^{2} + 23 = 0 .$ Approximate to the nearest hundredth.

Solution:

If we let $u = x^{2}$ , then $u^{2} = {(x^{2})}^{2} = x^{4}$ and we can write

$\begin{array}{l} x^{4} - 10 x^{2} + 23 = 0 \\ ↓ ↓ \\ u^{2} - 10 u + 23 = 0 \end{array}$

This equation does not factor; therefore, use the quadratic formula to find the solutions for u. Here $a = 1$ , $b = - 10$ , and $c = 23 .$

$\begin{array}{l} u & = & \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} \\ = & \frac{- (- 10) \pm \sqrt{{(- 10)}^{2} - 4 (1) (23)}}{2 (1)} \\ = & \frac{10 \pm \sqrt{8}}{2} \\ = & \frac{10 \pm 2 \sqrt{2}}{2} \\ = & 5 \pm \sqrt{2} \end{array}$

Therefore, $u = 5 \pm \sqrt{2} .$ Now back substitute $u = x^{2}$ and solve for x.

$\begin{array}{l} u & = & 5 - \sqrt{2} & or & u & = & 5 + \sqrt{2} \\ ↓ & ↓ \\ x^{2} & = & 5 - \sqrt{2} & x^{2} & = & 5 + \sqrt{2} \\ x & = & \pm \sqrt{5 - \sqrt{2}} & x & = & \pm \sqrt{5 + \sqrt{2}} \end{array}$

Round the four solutions as follows.

$\begin{array}{l} x & = & - \sqrt{5 - \sqrt{2}} \approx - 1.89 & x & = & - \sqrt{5 + \sqrt{2}} \approx - 2.53 \\ x & = & \sqrt{5 - \sqrt{2}} \approx 1.89 & x & = & \sqrt{5 + \sqrt{2}} \approx 2.53 \end{array}$

Answer: The solutions are approximately ±1.89, ±2.53.

If multiple roots and complex roots are counted, then the fundamental theorem of algebraIf multiple roots and complex roots are counted, then every polynomial with one variable will have as many roots as its degree. implies that every polynomial with one variable will have as many roots as its degree. For example, we expect $f (x) = x^{3} - 8$ to have three roots. In other words, the equation

$x^{3} - 8 = 0$ should have three solutions. To find them one might first think of trying to extract the cube roots just as we did with square roots,

$\begin{array}{l} x^{3} - 8 & = & 0 \\ x^{3} & = & 8 \\ x & = & \sqrt[3]{8} \\ x & = & 2 \end{array}$

As you can see, this leads to one solution, the real cube root. There should be two others; let’s try to find them.

Example 9

Find the set of all roots: $f (x) = x^{3} - 8 .$

Solution:

Notice that the expression $x^{3} - 8$ is a difference of cubes and recall that $a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2}) .$ Here $a = x$ and $b = 2$ and we can write

$\begin{matrix} x^{3} - 8 & = & 0 \\ (x - 2) (x^{2} + 2 x + 4) & = & 0 \end{matrix}$

Next apply the zero-product property and set each factor equal to zero. After setting the factors equal to zero we can then solve the resulting equation using the appropriate methods.

$\begin{array}{l} x - 2 & = & 0 & or & x^{2} + 2 x + 4 & = & 0 \\ x & = & 2 & x & = & \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} \\ = & \frac{- (2) \pm \sqrt{{(2)}^{2} - 4 (1) (4)}}{2 (1)} \\ = & \frac{- 2 \pm \sqrt{- 12}}{2} \\ = & \frac{- 2 \pm 2 i \sqrt{3}}{2} \\ = & - 1 \pm i \sqrt{3} \end{array}$

Using this method we were able to obtain the set of all three roots ${2, - 1 \pm i \sqrt{3}}$ , one real and two complex.

Answer: ${2, - 1 \pm i \sqrt{3}}$

Sometimes the roots of a function will occur multiple times. For example, $g (x) = {(x - 2)}^{3}$ has degree three where the roots can be found as follows:

$\begin{array}{l} \begin{array}{l} {(x - 2)}^{3} & = & 0 \\ (x - 2) (x - 2) (x - 2) & = & 0 \end{array} \\ \begin{array}{l} x - 2 & = & 0 & or & x - 2 & = & 0 & or & x - 2 & = & 0 \\ x & = & 2 & x & = & 2 & x & = & 2 \end{array} \end{array}$

Even though g is of degree 3 there is only one real root {2}; it occurs 3 times.

Key Takeaways

The quadratic formula can solve any quadratic equation. However, it is sometimes not the most efficient method.
If a quadratic equation can be solved by factoring or by extracting square roots you should use that method.
We can sometimes transform equations into equations that are quadratic in form by making an appropriate u-substitution. After solving the equivalent equation, back substitute and solve for the original variable.
Counting multiple and complex roots, the fundamental theorem of algebra guarantees as many roots as the degree of a polynomial equation with one variable.

Topic Exercises

Part A: Solving Quadratic Equations

Solve.

$x^{2} - 9 x = 0$
$x^{2} + 10 x = 0$
$15 x^{2} + 6 x = 0$
$36 x^{2} - 18 x = 0$
$x^{2} - 90 = 0$
$x^{2} + 48 = 0$
$2 x^{2} + 1 = 0$
$7 x^{2} - 1 = 0$
$6 x^{2} - 11 x + 4 = 0$
$9 x^{2} + 12 x - 5 = 0$
$x^{2} + x + 6 = 0$
$x^{2} + 2 x + 8 = 0$
$4 t^{2} + 28 t + 49 = 0$
$25 t^{2} - 20 t + 4 = 0$
$u^{2} - 4 u - 1 = 0$
$u^{2} - 2 u - 11 = 0$
$2 {(x + 2)}^{2} = 11 + 4 x - 2 x^{2}$
$(2 x + 1) (x - 3) + 2 x^{2} = 3 (x - 1)$
${(3 x + 2)}^{2} = 6 (2 x + 1)$
${(2 x - 3)}^{2} + 5 x^{2} = 4 (2 - 3 x)$
$4 {(3 x - 1)}^{2} - 5 = 0$
$9 {(2 x + 3)}^{2} - 2 = 0$

Part B: Solving Equations Quadratic in Form

Find all solutions.

$x^{4} + x^{2} - 72 = 0$
$x^{4} - 17 x^{2} - 18 = 0$
$x^{4} - 13 x^{2} + 36 = 0$
$4 x^{4} - 17 x^{2} + 4 = 0$
$x + 2 \sqrt{x} - 3 = 0$
$x - \sqrt{x} - 2 = 0$
$x - 5 \sqrt{x} + 6 = 0$
$x - 6 \sqrt{x} + 5 = 0$
$x^{2 / 3} + 5 x^{1 / 3} + 6 = 0$
$x^{2 / 3} - 2 x^{1 / 3} - 35 = 0$
$4 x^{2 / 3} - 4 x^{1 / 3} + 1 = 0$
$3 x^{2 / 3} - 2 x^{1 / 3} - 1 = 0$
$5 x^{- 2} + 9 x^{- 1} - 2 = 0$
$3 x^{- 2} + 8 x^{- 1} - 3 = 0$
$8 x^{- 2} + 14 x^{- 1} - 15 = 0$
$9 x^{- 2} - 24 x^{- 1} + 16 = 0$
${(\frac{x - 3}{x})}^{2} - 2 (\frac{x - 3}{x}) - 24 = 0$
${(\frac{2 x + 1}{x})}^{2} + 9 (\frac{2 x + 1}{x}) - 36 = 0$
$2 {(\frac{x}{x + 1})}^{2} - 5 (\frac{x}{x + 1}) - 3 = 0$
$3 {(\frac{x}{3 x - 1})}^{2} + 13 (\frac{x}{3 x - 1}) - 10 = 0$
$4 y^{- 2} - 9 = 0$
$16 y^{- 2} + 4 y^{- 1} = 0$
$30 y^{2 / 3} - 15 y^{1 / 3} = 0$
$y^{2 / 3} - 9 = 0$
$81 y^{4} - 1 = 0$
$5 {(\frac{1}{x + 2})}^{2} - 3 (\frac{1}{x + 2}) - 2 = 0$
$12 {(\frac{x}{2 x - 3})}^{2} - 11 (\frac{x}{2 x - 3}) + 2 = 0$
$10 x^{- 2} - 19 x^{- 1} - 2 = 0$
$x^{1 / 2} - 3 x^{1 / 4} + 2 = 0$
$x + 5 \sqrt{x} - 50 = 0$
$8 x^{2 / 3} + 7 x^{1 / 3} - 1 = 0$
$x^{4 / 3} - 13 x^{2 / 3} + 36 = 0$
$y^{4} - 14 y^{2} + 46 = 0$
$x^{4 / 3} - 2 x^{2 / 3} + 1 = 0$
$2 y^{- 2} - y^{- 1} - 1 = 0$
$2 x^{- 2 / 3} - 3 x^{- 1 / 3} - 2 = 0$
$4 x^{- 1} - 17 x^{- 1 / 2} + 4 = 0$
$3 x^{- 1} - 8 x^{- 1 / 2} + 4 = 0$
$2 x^{1 / 3} - 3 x^{1 / 6} + 1 = 0$
$x^{1 / 3} - x^{1 / 6} - 2 = 0$

Find all solutions. Round your answers to the nearest hundredth.

$x^{4} - 6 x^{2} + 7 = 0$
$x^{4} - 6 x^{2} + 6 = 0$
$x^{4} - 8 x^{2} + 14 = 0$
$x^{4} - 12 x^{2} + 31 = 0$
$4 x^{4} - 16 x^{2} + 13 = 0$
$9 x^{4} - 30 x^{2} + 1 = 0$

Find the set of all roots.

$f (x) = x^{3} - 1$
$g (x) = x^{3} + 1$
$f (x) = x^{3} - 27$
$g (x) = x^{4} - 16$
$h (x) = x^{4} - 1$
$h (x) = x^{6} - 1$
$f (x) = {(2 x - 1)}^{3}$
$g (x) = x^{2} {(x - 4)}^{2}$
$f (x) = x^{3} - q^{3}$ , $q > 0$
$f (x) = x^{3} + q^{3}$ , $q > 0$

Find all solutions.

$x^{6} + 7 x^{3} - 8 = 0$
$x^{6} - 7 x^{3} - 8 = 0$
$x^{6} + 28 x^{3} + 27 = 0$
$x^{6} + 16 x^{3} + 64 = 0$
$| x^{2} + 2 x - 5 | = 1$
$| x^{2} - 2 x - 3 | = 3$
$| 2 x^{2} - 5 | = 4$
$| 3 x^{2} - 9 x | = 6$

Find a quadratic function with integer coefficients and the given set of roots. (Hint: If $r_{1}$ and $r_{2}$ are roots, then $(x - r_{1}) (x - r_{2}) = 0 .$ )

${\pm 3 i}$
${\pm i \sqrt{5}}$
${\pm \sqrt{3}}$
${\pm 2 \sqrt{6}}$
${1 \pm \sqrt{3}}$
${2 \pm 3 \sqrt{2}}$
${1 \pm 6 i}$
${2 \pm 3 i}$

Part C: Discussion Board

On a note card, write out your strategy for solving a quadratic equation. Share your strategy on the discussion board.
Make up your own equation that is quadratic in form. Share it and the solution on the discussion board.

Answers

0, 9
$- \frac{2}{5}$ , 0
$\pm 3 \sqrt{10}$
$\pm \frac{\sqrt{2}}{2} i$
$\frac{1}{2}, \frac{4}{3}$
$- \frac{1}{2} \pm \frac{\sqrt{23}}{2} i$
$- \frac{7}{2}$
$2 \pm \sqrt{5}$
$- \frac{3}{2}, \frac{1}{2}$
$\pm \frac{\sqrt{2}}{3}$
$\frac{2 \pm \sqrt{5}}{6}$

$\pm 2 \sqrt{2}$ , $\pm 3 i$
±2, ±3
1
4, 9
−27, −8
$\frac{1}{8}$
$- \frac{1}{2}$ , 5
$- \frac{2}{5}, \frac{4}{3}$
$\pm \frac{3}{5}$
$- \frac{3}{2}, - \frac{1}{3}$
$\pm \frac{2}{3}$
0, $\frac{1}{8}$
$\pm \frac{1}{3}, \pm \frac{i}{3}$
$- \frac{3}{2}$ , 6
1, 16
−1, $\frac{1}{512}$
$\pm \sqrt{7 - \sqrt{3}}$ , $\pm \sqrt{7 + \sqrt{3}}$
−2, 1
$\frac{1}{16}$ , 16
$\frac{1}{64}$ , 1
±1.26, ±2.10
±1.61, ±2.33
±1.06, ±1.69
${1, - \frac{1}{2} \pm \frac{\sqrt{3}}{2} i}$
${3, - \frac{3}{2} \pm \frac{3 \sqrt{3}}{2} i}$
${\pm 1, \pm i}$
${\frac{1}{2}}$
${q, - \frac{q}{2} \pm \frac{q \sqrt{3}}{2} i}$
−2, 1, $1 \pm i \sqrt{3}$ , $- \frac{1}{2} \pm \frac{\sqrt{3}}{2} i$
−3, −1, $\frac{3}{2} \pm \frac{3 \sqrt{3}}{2} i$ , $\frac{1}{2} \pm \frac{\sqrt{3}}{2} i$
$- 1 \pm \sqrt{7}$ , $- 1 \pm \sqrt{5}$
$\pm \frac{\sqrt{2}}{2}$ , $\pm \frac{3 \sqrt{2}}{2}$
$f (x) = x^{2} + 9$
$f (x) = x^{2} - 3$
$f (x) = x^{2} - 2 x - 2$
$f (x) = x^{2} - 2 x + 37$

Answer may vary