1 2 2nd-degree equation and word problems-x

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Solving Equations by Factoring

Example A. a. Solve x2 – 2x – 3 = 0

An important factoring application is to find solutions of higher degree equations via the following steps.

1. To solve a polynomial equation, set one side of the equation to be 0 by moving all the terms the other side,

One side is already 0.

1. To solve a polynomial equation, set one side of the equation to be 0 by moving all the terms the other side, 2. factor the polynomial, if possible,

One side is already 0.

1. To solve a polynomial equation, set one side of the equation to be 0 by moving all the terms the other side,2. factor the polynomial, if possible,

One side is already 0. Factoring the trinomial we have that (x – 3)(x + 1) = 0

1. To solve a polynomial equation, set one side of the equation to be 0 by moving all the terms the other side, 2. factor the polynomial, if possible, 3. then extract an answer from each binomial factor.

One side is already 0. Factoring the trinomial we have that (x – 3)(x + 1) = 0

One side is already 0. Factoring the trinomial we have that (x – 3)(x + 1) = 0 There are two linear factors.

We may extract one answer from each.

One side is already 0. Factoring the trinomial we have that (x – 3)(x + 1) = 0Hence x – 3 = 0x = 3

There are two linear factors. We may extract one answer from each.

One side is already 0. Factoring the trinomial we have that (x – 3)(x + 1) = 0Hence x – 3 = 0 or x + 1 = 0 x = 3 or x = –1

b. Solve 2 = 2x2 – 3x

b. Solve 2 = 2x2 – 3xMake the left side 0 by moving the 2 so 0 = 2x2 – 3x – 2

b. Solve 2 = 2x2 – 3xMake the left side 0 by moving the 2 so 0 = 2x2 – 3x – 2 Factoring the trinomial yield 0 = (2x + 1)(x – 2),

b. Solve 2 = 2x2 – 3xMake the left side 0 by moving the 2 so 0 = 2x2 – 3x – 2 Factoring the trinomial yield 0 = (2x + 1)(x – 2), hence 2x + 1 = 0 so x = – ½, or x – 2 = 0 or x = 2.

Following are some applications of 2nd degree equations.

b. Solve 2 = 2x2 – 3xMake the left side 0 by moving the 2 so 0 = 2x2 – 3x – 2 Factoring the trinomial yield 0 = (2x + 1)(x – 2), hence 2x + 1 = 0 so x = – ½, or x – 2 = 0 or x = 2.

The simplest type of formulas leads to 2nd degree equations are the ones of the form AB = C.

2nd-Degree-Equation Word Problems

The simplest type of formulas leads to 2nd degree equations are the ones of the form AB = C. Specifically, if the two factors A and B are linear binomials then their product is a 2nd degree expression.

The simplest type of formulas leads to 2nd degree equations are the ones of the form AB = C. Specifically, if the two factors A and B are linear binomials then their product is a 2nd degree expression. Hence the problem of finding A and B when their product Cis known, is a 2nd degree equation.

Example A. We have two positive numbers.The larger number is 2 less than three times of the smaller one. The product of two numbers is 96. Find the two numbers.

Example A. We have two positive numbers.The larger number is 2 less than three times of the smaller one. The product of two numbers is 96. Find the two numbers.Let the smaller number be x.

Example A. We have two positive numbers.The larger number is 2 less than three times of the smaller one. The product of two numbers is 96. Find the two numbers.Let the smaller number be x. Hence the larger one is (3x – 2).

Example A. We have two positive numbers.The larger number is 2 less than three times of the smaller one. The product of two numbers is 96. Find the two numbers.Let the smaller number be x. Hence the larger one is (3x – 2). The product of them is 96 so x(3x – 2) = 96

Example A. We have two positive numbers.The larger number is 2 less than three times of the smaller one. The product of two numbers is 96. Find the two numbers.Let the smaller number be x. Hence the larger one is (3x – 2). The product of them is 96 so x(3x – 2) = 96 expand 3x2 – 2x = 96

Example A. We have two positive numbers.The larger number is 2 less than three times of the smaller one. The product of two numbers is 96. Find the two numbers.Let the smaller number be x. Hence the larger one is (3x – 2). The product of them is 96 so x(3x – 2) = 96 expand 3x2 – 2x = 963x2 – 2x – 96 = 0

make the right side 0

Example A. We have two positive numbers.The larger number is 2 less than three times of the smaller one. The product of two numbers is 96. Find the two numbers.Let the smaller number be x. Hence the larger one is (3x – 2). The product of them is 96 so x(3x – 2) = 96 expand 3x2 – 2x = 963x2 – 2x – 96 = 0(3x + 16)(x – 6) = 0So x = –16/3 or x = 6.

Example A. We have two positive numbers.The larger number is 2 less than three times of the smaller one. The product of two numbers is 96. Find the two numbers.Let the smaller number be x. Hence the larger one is (3x – 2). The product of them is 96 so x(3x – 2) = 96 expand 3x2 – 2x = 963x2 – 2x – 96 = 0. (3x + 16)(x – 6) = 0So x = –16/3 or x = 6. Hence the two numbers are 6 and 16.

Many physics formulas are 2nd degree. 2nd-Degree-Equation Word Problems

Many physics formulas are 2nd degree. If a stone is thrown straight up on Earth then

h = -16t2 + vt

whereh = height in feett = time in secondv = upward speed in feet per second

h = -16t2 + vt

height = -16t2 + vt after t secondswhere

h = height in feett = time in secondv = upward speed in feet per second

h = -16t2 + vt

h = height in feett = time in secondv = upward speed in feet per second Example B. If a stone is thrown straightup at a speed of 64 ft per second,a. how high is it after 1 second?

h = -16t2 + vt

h = height in feett = time in secondv = upward speed in feet per second Example B. If a stone is thrown straightup at a speed of 64 ft per second,a. how high is it after 1 second? t = 1, v = 64,

h = -16t2 + vt

h = height in feett = time in secondv = upward speed in feet per second Example B. If a stone is thrown straightup at a speed of 64 ft per second,a. how high is it after 1 second? t = 1, v = 64, so h = -16(1)2 + 64(1)

h = -16t2 + vt

h = height in feett = time in secondv = upward speed in feet per second Example B. If a stone is thrown straightup at a speed of 64 ft per second,a. how high is it after 1 second? t = 1, v = 64, so h = -16(1)2 + 64(1) h = -16 + 64 = 48 ft

h = -16t2 + vt

h = height in feett = time in secondv = upward speed in feet per second Example B. If a stone is thrown straightup at a speed of 64 ft per second,a. how high is it after 1 second? t = 1, v = 64, so h = -16(1)2 + 64(1) h = -16 + 64 = 48 ft

Hence after 1 second, the stone is 48 ft above the ground.

b. How long will it take for it to fall back to the ground?2nd-Degree-Equation Word Problems

b. How long will it take for it to fall back to the ground?To fall back to the ground means the height h is 0.

b. How long will it take for it to fall back to the ground?To fall back to the ground means the height h is 0. Hence h = 0, v = 64, need to find t.0 = -16 t2 + 64t

b. How long will it take for it to fall back to the ground?To fall back to the ground means the height h is 0. Hence h = 0, v = 64, need to find t.0 = -16 t2 + 64t 0 = -16t(t – 4)

b. How long will it take for it to fall back to the ground?To fall back to the ground means the height h is 0. Hence h = 0, v = 64, need to find t.0 = -16 t2 + 64t 0 = -16t(t – 4)t = 0

b. How long will it take for it to fall back to the ground?To fall back to the ground means the height h is 0. Hence h = 0, v = 64, need to find t.0 = -16 t2 + 64t 0 = -16t(t – 4)t = 0 or t – 4 = 0 or t = 4

b. How long will it take for it to fall back to the ground?To fall back to the ground means the height h is 0. Hence h = 0, v = 64, need to find t.0 = -16 t2 + 64t 0 = -16t(t – 4)t = 0 or t – 4 = 0 or t = 4 Therefore it takes 4 seconds.

c. What is the maximum height obtained?

c. What is the maximum height obtained?Since it takes 4 seconds for the stone to fall back to the ground, at 2 seconds it must reach the highest point.

c. What is the maximum height obtained?Since it takes 4 seconds for the stone to fall back to the ground, at 2 seconds it must reach the highest point. Hence t = 2, v = 64, need to find h.

c. What is the maximum height obtained?Since it takes 4 seconds for the stone to fall back to the ground, at 2 seconds it must reach the highest point. Hence t = 2, v = 64, need to find h.h = -16(2)2 + 64(2)

c. What is the maximum height obtained?Since it takes 4 seconds for the stone to fall back to the ground, at 2 seconds it must reach the highest point. Hence t = 2, v = 64, need to find h.h = -16(2)2 + 64(2)= - 64 + 128

c. What is the maximum height obtained?Since it takes 4 seconds for the stone to fall back to the ground, at 2 seconds it must reach the highest point. Hence t = 2, v = 64, need to find h.h = -16(2)2 + 64(2)= - 64 + 128= 64 (ft)

c. What is the maximum height obtained?Since it takes 4 seconds for the stone to fall back to the ground, at 2 seconds it must reach the highest point. Hence t = 2, v = 64, need to find h.h = -16(2)2 + 64(2)= - 64 + 128= 64 (ft)Therefore, the stone reached the maximum height of 64 feet.

2nd-Degree-Equation Word ProblemsFormulas of area in mathematics also lead to 2nd degree equations.

Area of a Rectangle

Formulas of area in mathematics also lead to 2nd degree equations.

Area of a RectangleGiven a rectangle, let L = length of a rectangle W = width of the rectangle,

Area of a RectangleGiven a rectangle, let L = length of a rectangle W = width of the rectangle, the area A of the rectangle is A = LW.

w A = LW

If L and W are in a given unit, then A is in unit2.

w A = LW

If L and W are in a given unit, then A is in unit2. For example: If L = 2 in, W = 3 in. then A = 2*3 = 6 in2

w A = LW

Example C. The length of a rectangle is 4 inches more than the width. The area is 21 in2. Find the length and width.

w A = LW

Example C. The length of a rectangle is 4 inches more than the width. The area is 21 in2. Find the length and width.Let x = width, then the length = (x + 4)

w A = LW

Example C. The length of a rectangle is 4 inches more than the width. The area is 21 in2. Find the length and width.Let x = width, then the length = (x + 4)LW = A, so (x + 4)x = 21

w A = LW

Example C. The length of a rectangle is 4 inches more than the width. The area is 21 in2. Find the length and width.Let x = width, then the length = (x + 4)LW = A, so (x + 4)x = 21 x2 + 4x = 21

w A = LW

Example C. The length of a rectangle is 4 inches more than the width. The area is 21 in2. Find the length and width.Let x = width, then the length = (x + 4)LW = A, so (x + 4)x = 21 x2 + 4x = 21 x2 + 4x – 21 = 0

w A = LW

Example C. The length of a rectangle is 4 inches more than the width. The area is 21 in2. Find the length and width.Let x = width, then the length = (x + 4)LW = A, so (x + 4)x = 21 x2 + 4x = 21 x2 + 4x – 21 = 0 (x + 7)(x – 3) = 0

w A = LW

Example C. The length of a rectangle is 4 inches more than the width. The area is 21 in2. Find the length and width.Let x = width, then the length = (x + 4)LW = A, so (x + 4)x = 21 x2 + 4x = 21 x2 + 4x – 21 = 0 (x + 7)(x – 3) = 0 x = - 7 or x = 3

w A = LW

Example C. The length of a rectangle is 4 inches more than the width. The area is 21 in2. Find the length and width.Let x = width, then the length = (x + 4)LW = A, so (x + 4)x = 21 x2 + 4x = 21 x2 + 4x – 21 = 0 (x + 7)(x – 3) = 0 x = - 7 or x = 3

w A = LW

Example C. The length of a rectangle is 4 inches more than the width. The area is 21 in2. Find the length and width.Let x = width, then the length = (x + 4)LW = A, so (x + 4)x = 21 x2 + 4x = 21 x2 + 4x – 21 = 0 (x + 7)(x – 3) = 0 x = - 7 or x = 3Therefore, the width is 3 and the length is 7.

w A = LW

Area of a Parallelogram

w A = LW

Area of a parallelogram A = BH

Area of a ParallelogramA parallelogram is the area enclosed by two sets of parallel lines.

w A = LW

Area of a ParallelogramA parallelogram is the area enclosed by two sets of parallel lines. If we move the shaded part as shown,

H=height

B=base

w A = LW

Area of a ParallelogramA parallelogram is the area enclosed by two sets of parallel lines. If we move the shaded part as shown,

H=height

B=base

w A = LW

Area of a ParallelogramA parallelogram is the area enclosed by two sets of parallel lines. If we move the shaded part as shown, we reshaped the parallelogram into a rectangle. B=base

H=height

w A = LW

Area of a ParallelogramA parallelogram is the area enclosed by two sets of parallel lines. If we move the shaded part as shown, we reshaped the parallelogram into a rectangle. So the area A of the parallelogram is A = BH

H=height

B=base

w A = LW

Example D. The area of the parallelogram shown is 27 ft2.Find x.

2x + 3

Example D. The area of the parallelogram shown is 27 ft2.Find x.The area is (base)(height) or thatx(2x + 3) = 27

2x + 3

x2x2 + 3x = 27

2x + 3

x2x2 + 3x = 27 2x2 + 3x – 27 = 0

2x + 3

x2x2 + 3x = 27 2x2 + 3x – 27 = 0 Factor (2x + 9)(x – 3) = 0

2x + 3

x2x2 + 3x = 27 2x2 + 3x – 27 = 0 Factor (2x + 9)(x – 3) = 0 x = –9/2, x = 3 ft

2x + 3

Area of a Triangle

2x + 3

Area of a triangle A = ½ BH

Area of a TriangleGiven the base (B) and the height (H) of a triangle as shown.

2x + 3

Take another copy and place it above the original one as shown .

2x + 3

Take another copy and place it above the original one as shown.

2x + 3

Take another copy and place it above the original one as shown.We obtain a parallelogram.

2x + 3

Take another copy and place it above the original one as shown.We obtain a parallelogram.If A is the area of the triangle,

2x + 3

Take another copy and place it above the original one as shown.We obtain a parallelogram.If A is the area of the triangle,

then 2A = HB or .A = BH 2 Area of a triangle A = ½ BH

Example E. The base of a triangle is 3 inches shorter than the twice of the height. The area is 10 in2. Find the base and height.

Example E. The base of a triangle is 3 inches shorter than the twice of the height. The area is 10 in2. Find the base and height.Let x = height, then the base = (2x – 3)

2x– 3

Example E. The base of a triangle is 3 inches shorter than the twice of the height. The area is 10 in2. Find the base and height.Let x = height, then the base = (2x – 3)Hence, use the formula 2A = BH 2*10 = (2x – 3) x

2x– 3

Example E. The base of a triangle is 3 inches shorter than the twice of the height. The area is 10 in2. Find the base and height.Let x = height, then the base = (2x – 3)Hence, use the formula 2A = BH 2*10 = (2x – 3) x20 = 2x2 – 3x

2x– 3

Example E. The base of a triangle is 3 inches shorter than the twice of the height. The area is 10 in2. Find the base and height.Let x = height, then the base = (2x – 3)Hence, use the formula 2A = BH 2*10 = (2x – 3) x20 = 2x2 – 3x0 = 2x2 – 3x – 20

2x– 3

Example E. The base of a triangle is 3 inches shorter than the twice of the height. The area is 10 in2. Find the base and height.Let x = height, then the base = (2x – 3)Hence, use the formula 2A = BH 2*10 = (2x – 3) x20 = 2x2 – 3x0 = 2x2 – 3x – 20 0 = (x – 4)(2x + 5)

2x– 3

Example E. The base of a triangle is 3 inches shorter than the twice of the height. The area is 10 in2. Find the base and height.Let x = height, then the base = (2x – 3)Hence, use the formula 2A = BH 2*10 = (2x – 3) x20 = 2x2 – 3x0 = 2x2 – 3x – 20 0 = (x – 4)(2x + 5)x = 4 or x = -5/2

2x– 3

Example E. The base of a triangle is 3 inches shorter than the twice of the height. The area is 10 in2. Find the base and height.Let x = height, then the base = (2x – 3)Hence, use the formula 2A = BH 2*10 = (2x – 3) x20 = 2x2 – 3x0 = 2x2 – 3x – 20 0 = (x – 4)(2x + 5)x = 4 or x = -5/2

2x– 3

Example E. The base of a triangle is 3 inches shorter than the twice of the height. The area is 10 in2. Find the base and height.Let x = height, then the base = (2x – 3)Hence, use the formula 2A = BH 2*10 = (2x – 3) x20 = 2x2 – 3x0 = 2x2 – 3x – 20 0 = (x – 4)(2x + 5)x = 4 or x = -5/2Therefore the height is 4 in. and the base is 5 in.

2x– 3

2nd-Degree-Equation Word ProblemsCoin Flips and Probability

2nd-Degree-Equation Word ProblemsCoin Flips and Probability We have a coin and we like to determine the “chances” that it would land on the head (H) or that tail (T) when it’s flipped.

We flip the coin 100 times and 50 times it landed H’s so we estimate the probability to get a H, or prob(H) = 50/100 = 0.5.

We have a coin and we like to determine the “chances” that it would land on the head (H) or that tail (T) when it’s flipped.

We flip the coin 100 times and 50 times it landed H’s so we estimate the probability to get a H, or prob(H) = 50/100 = 0.5. Note that it must have landed T’s 50 times so prob(T) = 0.5.

We flip the coin 100 times and 50 times it landed H’s so we estimate the probability to get a H, or prob(H) = 50/100 = 0.5. Note that it must have landed T’s 50 times so prob(T) = 0.5. We say this coin is fair because we’ve a 50–50 chances to obtain a head or tail on a flip.

However if 60 of the 100 flips are H’s, so 40 flips are T’s, then prob(H) = 60/100 = 0.6 and the prob(T) = 40/100 = 0.4.

However if 60 of the 100 flips are H’s, so 40 flips are T’s, then prob(H) = 60/100 = 0.6 and the prob(T) = 40/100 = 0.4. We say this coin is unfair because it’s not a 50–50 split.

However if 60 of the 100 flips are H’s, so 40 flips are T’s, then prob(H) = 60/100 = 0.6 and the prob(T) = 40/100 = 0.4. We say this coin is unfair because it’s not a 50–50 split.Since H and T are the only possible outcomes for coin flips,we must have that prob(H) + prob(T) = 1.

However if 60 of the 100 flips are H’s, so 40 flips are T’s, then prob(H) = 60/100 = 0.6 and the prob(T) = 40/100 = 0.4. We say this coin is unfair because it’s not a 50–50 split.Since H and T are the only possible outcomes for coin flips,we must have that prob(H) + prob(T) = 1. If prob(H) = p and prob(T) = q, then p + q = 1, so q = (1 – p). (Hence if we know one, we know the other.)

2nd-Degree-Equation Word ProblemsDouble Flips

We flip a coin twice and note the sides it landed on in order.

We flip a coin twice and note the sides it landed on in order. There are four possible outcomes: HH, HT, TH and TT.

Fact: The probability of a double–flip outcomes is the product of probabilities of each outcome*.

Fact: The probability of a double–flip outcomes is the product of probabilities of each outcome*. For example if prob(H) = 0.6 and prob(T) = 0.4,

then prob(HH) = (0.6)(0.6) = 0.36,

then prob(HH) = (0.6)(0.6) = 0.36, prob(TT) = (0.4) (0.4) = 0.16,

then prob(HH) = (0.6)(0.6) = 0.36, prob(TT) = (0.4) (0.4) = 0.16, and prob(HT) = prob(HT) = (0.6)(0.4) = 0.24

* We are assuming here that the 1st flip will not influence, the outcome of the 2nd flip (e.g. the 1st flip damaged the coin and altered the probabilities of future flips.)

* We are assuming here that the 1st flip will not influence, the outcome of the 2nd flip (e.g. the 1st flip damaged the coin and altered the probabilities of future flips.) If the result of an event A (flipping a coin) would not influence the result of another event B (flipping the coin 2nd time),we say the events A and B are independent.Here we are assuming the 1st flip is independent of the 2nd flip.

In general if prob(H) = p and prob(T) = q = (1 – p),

then prob(HH) = p * p = p2, prob(TT) = q * q = (1 – p)2 and that prob(HT) = prob(HT) = p * q = p(1 – p)

Double Flips

then prob(HH) = p * p = p2, prob(TT) = q * q = (1 – p)2 and that prob(HT) = prob(HT) = p * q = p(1 – p) Let’s use these facts to find the probability assignment of a coin.

Double Flips

Example F. We have a coin and we want to estimate the probability of getting a head when it’s flipped. Out of 100 double–flips, exactly 16 double flips are HT. Use this information to estimate p = prob(H).

Double Flips

Out of 100 double–flips, exactly 16 double flips are HTmeans that prob(HT) = 16/100 or 4/25.

Double Flips

Out of 100 double–flips, exactly 16 double flips are HTmeans that prob(HT) = 16/100 or 4/25. Since prob(HT) = pq = p(1 – p) = 4/25 so that p – p2 = 4/25.

Double Flips

Out of 100 double–flips, exactly 16 double flips are HTmeans that prob(HT) = 16/100 or 4/25. Since prob(HT) = pq = p(1 – p) = 4/25 so that p – p2 = 4/25.Multiple both sides by 25, we get 25p – 25p2 = 4.

Double Flips

Out of 100 double–flips, exactly 16 double flips are HTmeans that prob(HT) = 16/100 or 4/25. Since prob(HT) = pq = p(1 – p) = 4/25 so that p – p2 = 4/25.Multiple both sides by 25, we get 25p – 25p2 = 4. Hence 0 = 25p2 – 25p + 4 or 0 = (5p – 4)(5p – 1),

Double Flips

Out of 100 double–flips, exactly 16 double flips are HTmeans that prob(HT) = 16/100 or 4/25. Since prob(HT) = pq = p(1 – p) = 4/25 so that p – p2 = 4/25.Multiple both sides by 25, we get 25p – 25p2 = 4. Hence 0 = 25p2 – 25p + 4 or 0 = (5p – 4)(5p – 1),therefore prob(H) = p = 4/5 or p = 1/5.

Double Flips

Exercise A. Use the formula h = –16t2 + vt for the following problems.

1. A stone is thrown upward at a speed of v = 64 ft/sec,how long does it take for it’s height to reach 48 ft? 2. A stone is thrown upward at a speed of v = 64 ft/sec,how long does it take for it’s height to reach 28 ft? 3. A stone is thrown upward at a speed of v = 96 ft/sec,a. how long does it take for its height to reach 80 ft? Draw a picture. b. how long does it take for its height to reach the highest point? c. What is the maximum height it reached? 4. A stone is thrown upward at a speed of v = 128 ft/sec,a. how long does it take for its height to reach 256 ft? Draw a picture. How long does it take for its height to reach the highest point and what is the maximum height it reached?

B. Given the following area measurements, find x.2nd-Degree-Equation Word Problems

12 ft2

(x – 1)

8.12 ft2 x

(x + 4)

24 ft2 (3x – 1)

15 ft2x

18 ft2 x(4x + 1)

B. Given the following area measurements, find x.2nd-Degree-Equation Word Problems

2x + 1

2x – 3

2x + 1

x18km2

(x + 3)

(5x + 3)24km2

15. 16.16km2

35km22

2x – 1

1 2 2nd-degree equation and word problems-x

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