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Writing Equations from Solutions Solutions

Note

When given solutions, visualizing the equations they come from helps connect solutions to possible equations and their graphs on the coordinate plane. Remember to use technology to quickly visualize equations when needed.

To write a polynomial equation from given solutions, work backward.  
- Set each solution equal to the variable.  
- Then rewrite the solution as an equation set equal to zero .  
- Using the distributive property, multiply each expression set equal to zero together, resulting in one polynomial equation.
- The final equation will have integer coefficients.
 The Conjugate Root Theorem says: Polynomial equations with rational coefficients can have complex solutions, which will always come in conjugate pairs.
-  If $a + \sqrt{b}$ is an irrational root, then $a - \sqrt{b}$ is also a root.
-  If $a + b i$ is a complex root, then $a - b i$ is also a root.
 Important: The number of roots can help determine the degree of the equation; however, the number of roots can never be greater than the degree.

Note

There can be more than one equation for the given solution. If you arrive at another equation, use technology to check that the roots are the same. See the More to Explore activity for this lesson to compare equations using technology.

Example 1

Write a polynomial equation with integer coefficients using the given solutions. Classify the polynomial.

$x = - \frac{3}{2}, 5$

Note

Because there are two solutions, the result will be a second degree equation.

Implement

$\begin{matrix} x = - \frac{3}{2} & x = 5 \\ x + \frac{3}{2} = 0 & x - 5 = 0 \\ 2 x + 3 = 0 \end{matrix}$

$\begin{array}{rcl} (2 x + 3) (x - 5) & = & 0 \\ 2 x^{2} - 10 x + 3 x - 15 & = & 0 \end{array}$

$2 x^{2} - 7 x - 15 = 0$

Explain

Set every solution equal to x 
Rewrite each solution as an equation set equal to zero with integer coefficients 
Multiply the expressions together 
Write the polynomial equation set equal to zero

This is a quadratic trinomial .

Note

It is possible to write an equation using $(x + \frac{3}{2}) (x - 5) = 0$ to get the equation: $x^{2} - \frac{7}{2} x - \frac{15}{2} = 0$ . In this lesson, you are asked for integer coefficients due to the efficiency of solving. You can see more of this in the More to Explore activity for Lesson 23.

Example 2

Write a polynomial equation using the given solutions. Classify the polynomial.

$x = \pm \sqrt{3}, 6$

$\begin{matrix} x = - \sqrt{3} & x = \sqrt{3} & x = 6 \end{matrix}$

$\begin{matrix} x - (- \sqrt{3}) = 0 & x - \sqrt{3} = 0 & x - 6 = 0 \end{matrix} (x + \sqrt{3}) (x - \sqrt{3}) (x - 6) = 0 (x^{2} - 3) (x - 6) = 0 x^{3} - 6 x^{2} - 3 x + 18 = 0$

This is a cubic polynomial with four terms.

Note

The final equation will be a third degree polynomial because there are three solutions.

Example 3

Given the root $3 - 2 i$ , determine if there are any missing roots. Then write the quadratic equation with integer coefficients in standard form.

Implement

$3 - 2 i, 3 + 2 i x = 3 - 2 i, x = 3 + 2 i (x - (3 + 2 i)) (x - (3 - 2 i)) = 0$

$\begin{array}{rcl} (x - 3 - 2 i) (x - 3 + 2 i) & = & 0 \\ x^{2} - 3 x + 2 i x - 3 x + 9 - 6 i - 2 i x + 6 i - 4 i^{2} & = & 0 \\ x^{2} - 6 x + 9 - 4 (- 1) & = & 0 \\ x^{2} - 6 x + 9 + 4 & = & 0 \\ x^{2} - 6 x + 13 & = & 0 \end{array}$

Explain

Conjugate Root Theorem 
Set every solution equal to x
 Rewrite each solution as an equation set equal to zero 
Multiply the expressions together until the equation is simplified