Algebra 2 by Richard Wright
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Objectives:
SDA NAD Content Standards (2018): AII.4.1, AII.4.2, AII.5.1, AII.6.3
Consider a baseball that a batter hits into the air. A quadratic equation can be used to model the time the catcher has to get under the ball to catch it. A simpler way to solve a general quadratic equation than completing the square is by using the quadratic formula.
Think of the general form of a quadratic equation. Use completing the square to solve it symbolically.
ax2 + bx + c = 0
Add −c to both sides to get the x terms by themselves.
ax2 + bx = –c
Divide by a to get the leading coefficient to be 1.
$$ x^2+\frac{b}{a}x=-\frac{c}{a} $$
Add \(\left(\frac{b}{2}\right)^2\) to both sides. In this case \(b=\frac{b}{a}\), so \(\left(\frac{b}{2}\right)^2=\left(\frac{\frac{b}{a}}{2}\right)^2=\left(\frac{b}{2a}\right)^2\).
$$ x^2+\frac{b}{a}x+\left(\frac{b}{2a}\right)^2=\left(\frac{b}{2a}\right)^2-\frac{c}{a} $$
$$ x^2+\frac{b}{a}x+\frac{b^2}{4a^2}=\frac{b^2}{4a^2}-\frac{c}{a} $$
On the right side, get a common denominator to subtract the fractions.
$$ x^2+\frac{b}{a}x+\frac{b^2}{4a^2}=\frac{b^2}{4a^2}-\frac{4ac}{4a^2} $$
$$ x^2+\frac{b}{a}x+\frac{b^2}{4a^2}=\frac{b^2-4ac}{4a^2} $$
Write the left side as a perfect square.
$$ \left(x+\frac{b}{2a}\right)^2=\frac{b^2-4ac}{4a^2} $$
Square root both sides.
$$ \sqrt{\left(x+\frac{b}{2a}\right)^2}=\sqrt{\frac{b^2-4ac}{4a^2}} $$
$$ x+\frac{b}{2a}=\frac{\sqrt{b^2-4ac}}{2a} $$
Solve for x.
$$ x=-\frac{b}{2a}\pm\frac{\sqrt{b^2-4ac}}{2a} $$
$$ x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} $$
Since this is solved symbolically, then if the a, b, and c are known, they can be substituted into this equation to solve for x. This is called the quadratic formula.
Quadratic Formula
If ax2 + bx + c = 0, then
$$ x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} $$
To solve using the quadratic formula,
Solve x2 + 5x – 14 = 0 using the quadratic formula.
Solution
Put the equation in general form: ax2 + bx + c = 0. It already is.
a = 1, b = 5, c = –14
Substitute those into the quadratic formula.
$$ x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} $$
$$ x=\frac{-5\pm\sqrt{5^2-4\left(1\right)\left(-14\right)}}{2\left(1\right)} $$
Simplify the solutions.
$$ x=\frac{-5\pm\sqrt{81}}{2} $$
$$ x=\frac{-5\pm9}{2} $$
$$ x=\frac{-5+9}{2},\frac{-5-9}{2} $$
x = 2, –7
Solve 2x2 – 3x = 1 using the quadratic formula.
Solution
Put the equation in general form: ax2 + bx + c = 0.
2x2 – 3x – 1 = 0
a = 2, b = –3, c = –1
Substitute those into the quadratic formula.
$$ x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} $$
$$ x=\frac{3\pm\sqrt{\left(-3\right)^2-4\left(2\right)\left(-1\right)}}{2\left(2\right)} $$
Simplify the solutions.
$$ x=\frac{\mathbf{3}\pm\sqrt{\mathbf{17}}}{\mathbf{4}} $$
Solve 3x2 – 2x = -4 using the quadratic formula.
Solution
Put the equation in general form: ax2 + bx + c = 0.
3x2 – 2x + 4 = 0
a = 3, b = –2, c = 4
Substitute those into the quadratic formula.
$$ x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} $$
$$ x=\frac{2\pm\sqrt{\left(-2\right)^2-4\left(3\right)\left(4\right)}}{2\left(3\right)} $$
Simplify the solutions.
$$ x=\frac{2\pm\sqrt{-44}}{6} $$
$$ x=\frac{2\pm2\sqrt{11}i}{6} $$
Separate into standard from for complex numbers.
$$ x=\frac{2}{6}\pm\frac{2\sqrt{11}}{6}i $$
x = \(\frac{\mathbf{1}}{\mathbf{3}}\pm\frac{\sqrt{\mathbf{11}}}{\mathbf{3}}i\)
The part of the quadratic formula under the square root is called the discriminant. It can be used to classify the types of solutions a quadratic equation will have. For example, if the discriminant is negative, the solutions will include the square roots of negative numbers which means the solutions will be imaginary.
To use the discriminant,
Use the discriminant to classify the types of solutions of 2x2 − 5x + 10 = 0.
Solution
This is already in general form, so a = 2, b = −5, c = 10.
Substitute these into the discriminant: b2 − 4ac.
(–5)2 – 4(2)(10)
= 25 – 80
= –55
Since this is negative the quadratic equation has two imaginary solutions.
Use the discriminant to classify the types of solutions of \(-\frac{1}{2}x^2+\frac{3}{2}x+1=0\).
Solution
This is already in general form, so \(a=-\frac{1}{2}\), \(b=\frac{3}{2}\), c = 1.
Substitute these into the discriminant: b2 – 4ac.
$$ \left(\frac{3}{2}\right)^2-4\left(-\frac{1}{2}\right)\left(1\right) $$
$$ =\frac{9}{4}+\frac{4}{2} $$
$$ =\frac{9}{4}+\frac{8}{4} $$
$$ =\frac{17}{4} $$
Since this is positive the quadratic equation has two discrete real solutions.
If air resistance is ignored, a falling object’s height in meters, h, can be modeled by h = –4.9t2 + v0t + h0 where t is the time since it was dropped in seconds, v0 is the initial velocity of the object in meters per second, and h0 is the initial height of the object in meters. If a ball hit straight up at 45 m/s from a height of 1 m, how much time does a catcher have to move under to ball to catch it if he is 2.5 m tall?
Solution
Start by listing what is known.
v0 = 45, h0 = 1, h = 2.5
Substitute these into the given formula.
h = −4.9t2 + v0t + h0
2.5 = −4.9t2 + 45t + 1
This is a quadratic equation. Since both t2 and t appear, it could be solve by either factoring or the quadratic formula. Since there are decimal coefficients, starting with the quadratic formula might be fastest.
Put the equation in general form.
0 = −4.9t2 + 45t − 1.5
a = −4.9, b = 45, c = −1.5
Substitute those into the quadratic formula.
$$ x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} $$
$$ x=\frac{-45\pm\sqrt{\left(45\right)^2-4\left(-4.9\right)\left(-1.5\right)}}{2\left(-4.9\right)} $$
Simplify the solutions.
$$ x=\frac{-45\pm\sqrt{1995.6}}{-9.8} $$
$$ x=\frac{-45\pm44.67}{-9.8} $$
$$ x=\frac{-45+44.67}{-9.8},\frac{-45-44.67}{-9.8} $$
x = 0.03, 9.15
Interpret these answers. The catcher has either 0.03 s or 9.15 s to get under the ball. The 0.03 s is the time it takes the ball to get to 2.5 m height as the ball goes up. The 9.15 s is the time it takes for the ball to get to 2.5 m height as the ball comes back down. The second solution answers the question.
The catcher has 9.15 s to get under the ball to catch it.
123 #1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 23, 25, 27, 39, 61, Mixed Review = 20
Mixed Review