Mastery Check Solutions

Show What You Know

Note

Remember to write down your ideas and do NOT erase attempts. This way, you can see what is working and what is not and draw conclusions.

Use ${- 5, - 4, - 3, - 2, - 1, 0, 1, 2, 3, 4, 5}$ only once when completing parts A and B.

Write a quadratic equation that has two complex roots. Then find the roots.

$b^{2} - 4 a c < 0 b^{2} < 4 a c \begin{matrix} a = - 5, & b = 1, & c = - 2 \end{matrix} 1^{2} < 4 (- 5) (- 2) 1 < 40 - 5 x^{2} + 1 x - 2 = 0 x = \frac{- (1) \pm \sqrt{{(1)}^{2} - 4 (- 5) (- 2)}}{2 (- 5)} x = \frac{- 1 \pm \sqrt{1 - 40}}{- 10} = \frac{- 1 \pm \sqrt{- 39}}{- 10} x = \frac{- 1 \pm i \sqrt{39}}{- 10}$

$b^{2} - 4 a c < 0 b^{2} < 4 a c \begin{matrix} a = - 1, & b = 0, & c = - 2 \end{matrix} 0^{2} < 4 (- 1) (- 2) 0^{2} < 8 - 1 x^{2} + 0 x - 2 = 0 - x^{2} - 2 = 0 x = \frac{- (0) \pm \sqrt{{(0)}^{2} - 4 (- 1) (- 2)}}{2 (- 1)} x = \frac{0 \pm \sqrt{0 - 8}}{- 2} = \frac{\pm \sqrt{- 8}}{- 2} x = \frac{\pm 2 i \sqrt{2}}{- 2} = \frac{- 2 (\pm i \sqrt{2})}{- 2} x = \pm i \sqrt{2}$

$b^{2} - 4 a c < 0 b^{2} < 4 a c \begin{matrix} a = - 3, & b = 5, & c = - 5 \end{matrix} 5^{2} < 4 (- 3) (- 5) 25 < 60 - 3 x^{2} + 5 x - 5 = 0 x = \frac{- (5) \pm \sqrt{{(5)}^{2} - 4 (- 3) (- 5)}}{2 (- 3)} x = \frac{- 5 \pm \sqrt{25 - 60}}{- 6} = \frac{- 5 \pm \sqrt{- 35}}{- 6} x = \frac{- 5 \pm i \sqrt{35}}{- 6}$

Note

The value of $a \neq 0,$ for any quadratic equation. There are many potential equations, but all must have a discriminant that is less than zero.

Q: What formula can be used to determine the type of roots for a quadratic equation?

A: The discriminant

Q: What are the ways that you can solve for roots in a quadratic equation?

A: Factoring, completing the square, quadratic formula

See the More to Explore for this lesson to check your work using technology.

Write a quadratic equation that has two real, irrational roots. Then find the roots. Remember that any value that is used in part A cannot be used again in part B.

$b^{2} - 4 a c > 0 b^{2} > 4 a c \begin{matrix} a = 2, & b = 5, & c = - 1 \end{matrix} 5^{2} > 4 (2) (- 1) 25 > - 8 2 x^{2} + 5 x - 1 = 0 x = \frac{- (5) \pm \sqrt{{(5)}^{2} - 4 (2) (- 1)}}{2 (2)} x = \frac{- 5 \pm \sqrt{25 + 8}}{4} x = \frac{- 5 \pm \sqrt{33}}{4}$

$b^{2} - 4 a c > 0 b^{2} > 4 a c \begin{matrix} a = 1, & b = 4, & c = - 3 \end{matrix} 4^{2} > 4 (1) (- 3) 16 > - 12 1 x^{2} + 4 x - 3 = 0 x^{2} + 4 x - 3 = 0 x = \frac{- (4) \pm \sqrt{{(4)}^{2} - 4 (1) (- 3)}}{2 (1)} x = \frac{- 4 \pm \sqrt{16 + 12}}{2 (1)} = \frac{- 4 \pm \sqrt{28}}{2} x = \frac{- 4 \pm 2 \sqrt{7}}{2} = \frac{2 (- 2 \pm \sqrt{7})}{2} x = - 2 \pm \sqrt{7}$

$b^{2} - 4 a c > 0 b^{2} > 4 a c \begin{matrix} a = 1, & b = - 4, & c = - 2 \end{matrix} {(- 4)}^{2} > 4 (1) (- 2) 16 > - 8 1 x^{2} - 4 x - 2 = 0 x^{2} + 4 x - 2 = 0 x = \frac{- (- 4) \pm \sqrt{{(- 4)}^{2} - 4 (1) (- 2)}}{2 (1)} x = \frac{4 \pm \sqrt{16 + 8}}{2 (1)} = \frac{4 \pm \sqrt{24}}{2} x = \frac{4 \pm 2 \sqrt{6}}{2} = \frac{2 (2 \pm \sqrt{6})}{2} x = 2 \pm \sqrt{6}$

Note

Remember that any value that is used in part A cannot be used again in part B.

There are many potential equations, but all must have a discriminant that is greater than zero.

Explain how you know what type of solutions a quadratic equation will have.

The value of $a \neq 0$ , for any quadratic equation. Any quadratic equation with complex roots will have $b^{2} - 4 a c < 0$ , or $b^{2} < 4 a c$ . Any quadratic equation with real roots will have $b^{2} - 4 a c > 0$ , or $b^{2} > 4 a c$ .

Say What You Know

In your own words, talk about what you have learned using the objectives for this lesson and your work on this page.

Note

Restate the objectives of the lesson in your own words. If you are unable to restate the lesson objectives, go back and reread the objectives and then explain them.

Solve quadratic equations using the quadratic formula.
Determine the number of real or complex solutions by solving for the discriminant.