Test 13 (Lessons 25–26): The Quadratic Formula, Distance Formula, and Midpoint Formula Solutions

Solve using the Quadratic Formula. Show all work.
$2 x^{2} = 3 x + 9$

$2 x^{2} - 3 x - 9 = 0 a = 2, b = - 3, c = - 9 x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} = \frac{- (- 3) \pm \sqrt{{(- 3)}^{2} - 4 (2) (- 9)}}{2 (2)} x = \frac{3 \pm \sqrt{9 + 72}}{4} = \frac{3 \pm \sqrt{81}}{4} = \frac{3 \pm 9}{4} x = - \frac{6}{4}, \frac{12}{4}$

x = –1.5, 3

Calculate the distance between the y-intercept and the positive root using the equation from problem 1.

$y ‐ intercept : (0, - 9), postive x ‐ intercept : (3, 0) d = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}} d = \sqrt{{(0 - 3)}^{2} + {(- 9 - 0)}^{2}} d = 9.486$

d = 9.49

Fill in the blanks.

The midpoint formula calculates the point that is equidistant from either endpoint.
When the discriminant is in the form $b^{2} - 4 a c < 0,$ the roots will be complex .

Use the graph to answer problems 5–6.

Determine the midpoint of each side of triangle AGL.

$A (- 3, 3), G (3, 1) (\frac{- 3 + 3}{2}, \frac{3 + 1}{2}) (0, 2) G (3, 1), L (- 1, - 7) (\frac{3 + - 1}{2}, \frac{1 + - 7}{2}) (1, - 3) A (- 3, 3), L (- 1, - 7) (\frac{- 3 + - 1}{2}, \frac{3 + - 7}{2}) (- 2, - 2)$

$midpoints of the sides A G (0, 2), G L (1, - 3), A L (- 2, - 2)$

Connect all of the midpoints to form a new triangle. Find the perimeter of the new triangle.

$(0, 2) to (1, - 3) (0, 2) to (- 2, - 2) d = \sqrt{{(0 - 1)}^{2} + (2 - {(- 3))}^{2}} d = \sqrt{(0 - {(- 2))}^{2} + (2 - {(- 2))}^{2}} d = 5.099 d = 4.472 d = 5.10 d = 4.47 (- 2, - 2) t o (1, - 3) d = \sqrt{{(- 2 - 1)}^{2} + (- 2 - {(- 3))}^{2}} d = 3.162 d = 3.16$

$Perimeter = 5.10 + 4.47 + 3.16 P = 12.73$

The perimeter of the new triangle is 12.73 units.

Fill in the blanks.

The distance formula can be used to calculate the units between two ordered pairs.
The quadratic formula can be used to determine the roots (or solutions, x-intercepts) of a quadratic equation.

A slow-pitch softball pitcher releases the ball 2.25 feet above the ground with an initial velocity of 50 ft/s. If the batter hits the ball 2 feet above the ground, how long did it take for the ball to reach the batter? Recall $h = \frac{1}{2} a t^{2} + v t + s$ and $a = - 32 .$

$h = 2, a = - 32, v = 50, s = 2.25 2 = \frac{1}{2} (- 32) t^{2} + 50 t + 2.25 0 = - 16 t^{2} + 50 t + 0.25 a = - 16, b = 50, c = 0.25 x = \frac{- (50) \pm \sqrt{{(50)}^{2} - 4 (- 16) (0.25)}}{2 (- 16)} = \frac{- 50 \pm \sqrt{2500 + 16}}{- 32} x = - 0.005, 3.130$

The batter will hit the ball 3.13 seconds after it is pitched.

Determine the type of roots using the discriminant. Explain.
$1.5 x^{2} + 4 x = 0.2$

$1.5 x^{2} + 4 x - 0.2 = 0 a = 1.5, b = 4, c = - 0.2 b^{2} - 4 a c {(4)}^{2} - 4 (1.5) (- 0.2) 16 + 1.2 = 17.2$

Sample: The quadratic equation will have two real, irrational roots because the discriminant is greater than zero and not a perfect square.