Practice 1 Solutions

Be sure to carefully read the directions. Some problems have all solutions given. Others require including the conjugate before writing the equation.

Given the root(s), determine if there are any missing roots and write an equation of the polynomial in standard form with integer coefficients.

$x = 3, - 2$

$x - 3 = 0, x + 2 = 0 (x - 3) (x + 2) = 0 x^{2} + 2 x - 3 x - 6 = 0$

$x^{2} - x - 6 = 0$

$x = 5 i$

$\begin{matrix} x = - 5 i, & x = 5 i \\ x = (- 5 i) = 0, & x - 5 i = 0 \end{matrix} (x - (- 5 i)) (x - 5 i) = 0 (x + 5 i) (x - 5 i) = 0 x^{2} - 5 i x + 5 i x - 25 i^{2} = 0 x^{2} - 25 i^{2} = 0 x^{2} - 25 (- 1) = 0$

$x^{2} + 25 = 0$

$x = 1 - \sqrt{5}$

$\begin{array}{clcccccc} x = 1 - \sqrt{5}, & x = 1 + \sqrt{5} \\ x - (1 - \sqrt{5}) = 0, & x - (1 + \sqrt{5}) = 0 \end{array} (x - (1 + \sqrt{5})) (x - (1 - \sqrt{5})) = 0 (x - 1 - \sqrt{5}) (x - 1 + \sqrt{5}) = 0 x^{2} - x + x \sqrt{5} - x + 1 - \sqrt{5} - x \sqrt{5} + \sqrt{5} - \sqrt{25} = 0 x^{2} - 2 x + 1 - 5 = 0$

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

$x = - i, \sqrt{6}$

$\begin{array}{cccr} x = - i, & x = i, & x = - \sqrt{6}, & x = \sqrt{6} \\ x - (- i) = 0, & x - i = 0, & x - (- \sqrt{6}) = 0, & x - \sqrt{6} = 0 \end{array} (x - (- i)) (x - i) (x - (- \sqrt{6})) (x - \sqrt{6}) = 0 (x + i) (x - i) (x + \sqrt{6}) (x - \sqrt{6}) = 0 (x^{2} - i x + i x - i^{2}) (x^{2} - x \sqrt{6} + x \sqrt{6} - \sqrt{36}) = 0 (x^{2} - i^{2}) (x^{2} - 6) = 0 (x^{2} - (- 1)) (x^{2} - 6) = 0 (x^{2} + 1) (x^{2} - 6) = 0 x^{4} - 6 x^{2} + x^{2} - 6 = 0$

$x^{4} - 5 x^{2} - 6 = 0$

$x = \frac{1}{2}, \frac{2}{3}$

$\begin{array}{cl} x - \frac{1}{2} = 0, & x - \frac{2}{3} = 0 \end{array} (x - \frac{1}{2}) (x - \frac{2}{3}) = 0 (2 x - 1) (3 x - 2) = 0 6 x^{2} - 4 x - 3 x + 2 = 0$

$6 x^{2} - 7 x + 2 = 0$

Note

This problem is also shown in the More to Explore activity for Lesson 23.

$x = \pm \sqrt{11}, \frac{3}{4}$

$x - (- \sqrt{11}) = 0, x - \sqrt{11} = 0, x - \frac{3}{4} = 0 (x - (- \sqrt{11})) (x - \sqrt{11}) (x - \frac{3}{4}) = 0 (x + \sqrt{11}) (x - \sqrt{11}) (4 x - 3) = 0 (x^{2} - x \sqrt{11} + x \sqrt{11} - \sqrt{121}) (4 x - 3) = 0 (x^{2} - 11) (4 x - 3) = 0$

$4 x^{3} - 3 x^{2} - 44 x + 33 = 0$

$x = 0, 9$

$x = 0, x - 9 = 0 x (x - 9) = 0$

$x^{2} - 9 x = 0$

Solve.

$x^{3} - 7 x^{2} + 9 x - 63 = 0$

$\begin{array}{rcl} (x^{3} - 7 x^{2}) + (9 x - 63) & = & 0 \\ x^{2} (x - 7) + 9 (x - 7) & = & 0 \\ (x^{2} + 9) (x - 7) & = & 0 \\ (x^{2} - (- 9)) (x - 7) & = & 0 \\ (x + 3 i) (x - 3 i) (x - 7) & = & 0 \end{array}$

$x = \pm 3 i, 7$

Note

Q: How is the degree related to the number of solutions?

A: The degree determines the number of solutions you will solve for.

$9 x^{2} - 6 x \sqrt{2} + 2 = 0$

$\begin{array}{rcl} (3 x - \sqrt{2}) (3 x - \sqrt{2}) & = & 0 \\ 3 x - \sqrt{2} & = & 0 \\ 3 x & = & \sqrt{2} \end{array}$

$x = \frac{\sqrt{2}}{3}$

Note

This is an example of a repeated root.

The Conjugate Root Theorem is not necessary here because the middle coefficient is irrational.

$x^{2} - \frac{7}{25} = 0$

$\begin{array}{rcl} x^{2} & = & \frac{7}{25} \\ \sqrt{x^{2}} & = & \pm \sqrt{\frac{7}{25}} \\ x & = & \pm \frac{\sqrt{7}}{\sqrt{25}} \end{array}$

$x = \pm \frac{\sqrt{7}}{5}$

$3 x^{2} + 54 = 0$

$\begin{array}{rcl} 3 x^{2} & = & - 54 \\ x^{2} & = & - 18 \\ \sqrt{x^{2}} & = & \pm \sqrt{- 18} \\ x & = & \pm i \sqrt{18} \end{array}$

$x = \pm 3 i \sqrt{2}$

Note

This problem can be solved using the Square Root Property or by factoring.

$5 x^{3} - 15 x^{2} - 10 x + 30 = 0$

$\begin{array}{rcl} 5 (x^{3} - 3 x^{2} - 2 x + 6) & = & 0 \\ 5 (x^{2} (x - 3) - 2 (x - 3)) & = & 0 \\ 5 (x^{2} - 2) (x - 3) & = & 0 \\ 5 (x + \sqrt{2}) (x - \sqrt{2}) (x - 3) & = & 0 \end{array}$

$x = \pm \sqrt{2}, 3$

$x^{3} - 5 x^{2} - 6 x + 30 = 0$

$\begin{array}{rcl} (x^{3} - 5 x^{2}) + (- 6 x + 30) & = & 0 \\ x^{2} (x - 5) - 6 (x - 5) & = & 0 \\ (x^{2} - 6) (x - 5) & = & 0 \\ (x + \sqrt{6}) (x - \sqrt{6}) (x - 5) & = & 0 \end{array}$

$x = \pm \sqrt{6}, 5$

$x^{2} + 4 = 0$

$\begin{array}{rcl} x^{2} & = & - 4 \\ \sqrt{x^{2}} & = & \pm \sqrt{- 4} \\ x & = & \pm i \sqrt{4} \end{array}$

$x = \pm 2 i$

$x^{4} + 13 x^{2} + 36 = 0$

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

$x = \pm 3 i, \pm 2 i$

Note

Students may also solve by setting each parenthesis equal to 0 and taking the square root of each side.

${(5 x - 1)}^{2} = 4$

$\begin{array}{rcl} \sqrt{{(5 x - 1)}^{2}} & = & \pm \sqrt{4} \\ 5 x - 1 & = & \pm 2 \\ 5 x & = & 1 \pm 2 \\ x & = & \frac{1 \pm 2}{5} \end{array}$

$x = - \frac{1}{5}, \frac{3}{5}$

$2 {(x - 5)}^{2} = - 48$

$\begin{array}{rcl} {(x - 5)}^{2} & = & - 24 \\ \sqrt{{(x - 5)}^{2}} & = & \pm \sqrt{- 24} \\ x - 5 & = & \pm i \sqrt{24} \\ x - 5 & = & \pm 2 i \sqrt{6} \end{array}$

$x = 5 \pm 2 i \sqrt{6}$

$x^{3} - 7 x^{2} + 8 x - 56 = 0$

$\begin{array}{rcl} (x^{3} - 7 x^{2}) + (8 x - 56) & = & 0 \\ x^{2} (x - 7) + 8 (x - 7) & = & 0 \\ (x^{2} + 8) (x - 7) & = & 0 \\ (x^{2} - (- 8)) (x - 7) & = & 0 \\ (x + i \sqrt{8}) (x - i \sqrt{8}) (x - 7) & = & 0 \\ (x + 2 i \sqrt{2}) (x - 2 i \sqrt{2}) (x - 7) & = & 0 \end{array}$

$x = \pm 2 i \sqrt{2}, 7$

$x^{2} - 10 = 0$

$\begin{array}{rcl} x^{2} & = & 10 \\ \sqrt{x^{2}} & = & \pm \sqrt{10} \end{array}$

$x = \pm \sqrt{10}$

${(x + 8)}^{2} = - 9$

$\begin{array}{rcl} \sqrt{{(x + 8)}^{2}} & = & \pm \sqrt{- 9} \\ x + 8 & = & \pm 3 i \end{array}$

$x = - 8 \pm 3 i$

$x^{4} + 6 x^{2} - 7 = 0$

$\begin{array}{rcl} (x^{2} + 7) (x^{2} - 1) & = & 0 \\ (x^{2} - (- 7)) (x^{2} - 1) & = & 0 \\ (x + i \sqrt{7}) (x - i \sqrt{7}) (x + 1) (x - 1) & = & 0 \end{array}$

$x = \pm i \sqrt{7}, \pm 1$

$x^{2} - x \sqrt{13} = 0$

$x (x - \sqrt{13}) = 0$

$x = 0, \sqrt{13}$

Note

The Conjugate Root Theorem is not necessary here because a coefficient is irrational.