Practice 2 Solutions

Solve. Write all answers in simplest form.

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

$a = 1, b = - 3, c = 9 x = \frac{- (- 3) \pm \sqrt{{(- 3)}^{2} - 4 (1) (9)}}{2 (1)} x = \frac{3 \pm \sqrt{9 - (36)}}{2} x = \frac{3 \pm \sqrt{- 27}}{2} x = \frac{3 \pm i \sqrt{27}}{2}$

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

$5 x^{2} + 2 x = 3 x^{2} + 11$

$2 x^{2} + 2 x - 11 = 0 a = 2, b = 2, c = - 11 x = \frac{- (2) \pm \sqrt{{(2)}^{2} - 4 (2) (- 11)}}{2 (2)} x = \frac{- 2 \pm \sqrt{4 - (- 88)}}{4} = \frac{- 2 \pm \sqrt{4 + 88}}{4} x = \frac{- 2 \pm \sqrt{92}}{4} = \frac{- 2 \pm 2 \sqrt{23}}{4} x = \frac{2 (- 1 \pm \sqrt{23})}{2 (2)}$

$x = \frac{- 1 \pm \sqrt{23}}{2}$

$4 x^{2} - 12 x + 9 = 0$

$a = 4, b = - 12, c = 9 x = \frac{- (- 12) \pm \sqrt{{(- 12)}^{2} - 4 (4) (9)}}{2 (4)} x = \frac{12 \pm \sqrt{144 - 144}}{8} x = \frac{12 \pm \sqrt{0}}{8} x = \frac{12}{8}$

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

Note

This equation is the perfect square trinomial ${(3 x - 2)}^{2}$ .

$- x^{2} - 2 = 5 x + 7$

$- x^{2} - 5 x - 9 = 0 a = - 1, b = - 5, c = - 9 x = \frac{- (- 5) \pm \sqrt{{(- 5)}^{2} - 4 (- 1) (- 9)}}{2 (- 1)} x = \frac{5 \pm \sqrt{25 - (36)}}{- 2} x = \frac{5 \pm \sqrt{- 11}}{- 2}$

$x = \frac{5 \pm i \sqrt{11}}{- 2} or \frac{- 5 \pm i \sqrt{11}}{2}$

$x^{2} + 6 x + 13 = 0$

$a = 1, b = 6, c = 13 x = \frac{- (6) \pm \sqrt{{(6)}^{2} - 4 (1) (13)}}{2 (1)} x = \frac{- 6 \pm \sqrt{36 - 52}}{2} x = \frac{- 6 \pm \sqrt{- 16}}{2} x = \frac{- 6 \pm i \sqrt{16}}{2} x = \frac{- 6 \pm 4 i}{2} x = \frac{2 (- 3 \pm 2 i)}{2}$

$x = - 3 \pm 2 i$

$7 x^{2} + 8 x = 5 x^{2} + 6 x$

$2 x^{2} + 2 x = 0 a = 2, b = 2, c = 0 x = \frac{- (2) \pm \sqrt{{(2)}^{2} - 4 (2) (0)}}{2 (2)} x = \frac{- 2 \pm \sqrt{4 - 0}}{4} x = \frac{- 2 \pm \sqrt{4}}{4} x = \frac{- 2 \pm 2}{4} x = - \frac{4}{4}, \frac{0}{4}$

$x = - 1, 0$

Note

If you notice this equation can be factored, try that as well and see if you solve for the same solutions.

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

$a = - 3, b = 1, c = 1 x = \frac{- (1) \pm \sqrt{{(1)}^{2} - 4 (- 3) (1)}}{2 (- 3)} x = \frac{- 1 \pm \sqrt{1 - (- 12)}}{- 6} x = \frac{- 1 \pm \sqrt{1 + 12}}{- 6} x = \frac{- 1 \pm \sqrt{13}}{- 6}$

$x = \frac{- 1 \pm \sqrt{13}}{- 6} or \frac{1 \pm \sqrt{13}}{6}$

$4 x^{2} + 25 = 0$

$a = 4, b = 0, c = 25 x = \frac{- (0) \pm \sqrt{{(0)}^{2} - 4 (4) (25)}}{2 (4)} x = \frac{\pm \sqrt{- 400}}{8} x = \frac{\pm i \sqrt{400}}{8} x = \pm \frac{20 i}{8}$

$x = \pm \frac{5 i}{2}$

Note

You may choose to solve this problem using the methods learned in previous lessons by isolating $x^{2}$ and taking the square root of both sides.

Determine the type of roots to the quadratic equation using the discriminant. Explain.

$6 x^{2} - 2 = 3 x$

$6 x^{2} - 3 x - 2 = 0 a = 6, b = - 3, c = - 2 b^{2} - 4 a c {(- 3)}^{2} - 4 (6) (- 2) 9 - (- 48) 9 + 48 57$

This equation has two real, irrational solutions.

$x^{2} + x + 5 = 0$

$a = 1, b = 1, c = 5 b^{2} - 4 a c {(1)}^{2} - 4 (1) (5) 1 - 20 - 19$

This equation has two complex solutions.

$x^{2} - 2 x = - 1$

$x^{2} - 2 x + 1 = 0 a = 1, b = - 2, c = 1 b^{2} - 4 a c {(- 2)}^{2} - 4 (1) (1) 4 - 4 0$

This equation has one real, rational solution.

Note

When the discriminant is equal to zero, it can also be called a double root.

$x^{2} + 5 x + 6 = 0$

$a = 1, b = 5, c = 6 b^{2} - 4 a c {(5)}^{2} - 4 (1) (6) 25 - 24 1$

This equation has two real, rational solutions.

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

$- x^{2} + 10 x - 3 = 0 a = - 1, b = 10, c = - 3 b^{2} - 4 a c {(10)}^{2} - 4 (- 1) (- 3) 100 - 12 88$

This equation has two real, irrational solutions.

$6 x^{2} + 15 = 0$

$a = 6, b = 0, c = 15 b^{2} - 4 a c {(0)}^{2} - 4 (6) (15) - 360$

This equation has two complex solutions.

$7 x^{2} + 13 x - 2 = 0$

$a = 7, b = 13, c = - 2 b^{2} - 4 a c {(13)}^{2} - 4 (7) (- 2) 169 + 56 225$

This equation has two real, rational solutions

$10 x^{2} - 3 x = 9 x^{2} + 5$

$x^{2} - 3 x - 5 = 0 a = 1, b = - 3, c = - 5 b^{2} - 4 a c {(- 3)}^{2} - 4 (1) (- 5) 9 - (- 20) 9 + 20 29$

This equation has two real, irrational solutions.

A rock is pushed off a 45 meter cliff at an initial velocity of 0 meters per second. Find the time it takes to hit the ground in meters per second.

$\begin{array}{rcl} 0 & = & \frac{1}{2} (- 9.8) t^{2} + 0 t + 45 \\ 0 & = & - 4.9 t^{2} + 45 \\ 4.9 t^{2} & = & 45 \\ t^{2} & = & 9.18 \\ t & = & - 3.03, 3.03 \end{array}$

It takes 3.03 seconds to hit the ground.

Note

You could choose to use the quadratic formula; however, this method is more efficient.

A ball is thrown with an initial velocity of 42 feet per second from a starting height of 9 feet. How long until the ball is 2 feet from the ground? Round your answer to the nearest hundredth.

$2 = \frac{1}{2} (- 32) t^{2} + 42 t + 9 0 = - 16 t^{2} + 42 t + 7 a = - 16, b = 42, c = 7 t = \frac{- 42 \pm \sqrt{42^{2} - 4 (- 16) (7)}}{2 (- 16)} t = \frac{- 42 \pm \sqrt{2212}}{- 32} t = \frac{- 42 \pm 47.03}{- 32} t = - 0.16, 2.78$

It took 2.78 seconds for the rock to be 2 feet from the ground.

Note

The negative time value is extraneous since we cannot go back in time.

Elena jumped from the high dive into the local pool. She left the springboard at a velocity of 10 feet per second at a height of 15 feet. How long did it take until she touched the water? Round your answer to the nearest hundredth.

$- \frac{1}{2} (32) t^{2} + 10 t + 15 = 0 - 16 t^{2} + 10 t + 15 = 0 a = - 16, b = 10, c = 15 t = \frac{- 10 \pm \sqrt{{(10)}^{2} - 4 (- 16) (15)}}{2 (- 16)} t = \frac{- 10 \pm \sqrt{100 + 960}}{- 32} t = \frac{- 10 \pm 32.56}{- 32} t = - 0.71, 1.33$

Elena entered the water at 1.33 seconds.

Kristin and Russell were playing catch in the back yard. The ball traveled at a velocity of 65 feet per second at a height of 5.5 feet. If Russell catches the ball 5 feet off the ground, how long after Kristin threw the ball did Russell catch it?

$- \frac{1}{2} (32) t^{2} + 65 t + 5.5 = 5 - 16 t^{2} + 65 t + 0.5 = 0 a = - 16, b = 65, c = 0.5 t = \frac{- 65 \pm \sqrt{{(65)}^{2} - 4 (- 16) (0.5)}}{2 (- 16)} t = \frac{- 65 \pm \sqrt{4225 + 16}}{- 32} t = \frac{- 65 \pm 65.12}{- 32} t = - 0.003, 4.07$

Russell caught the ball after 4.07 seconds.