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20 Quadratic Equation Questions : Word problems and Complex Roots

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(Full Problem Set found in attachment)

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7 questions on Quadratic Equations by Factoring:

1) Using the factoring method, solve for the roots of each equation. Place equation in standard form before factoring. Check your solutions and show the check.
3x2 = 75

2) Using the factoring method, solve for the roots of each equation. Place equation in standard form before factoring. Check your solutions and show the check.
4x2 + 3x = 4x

3) Using the factoring method, solve for the roots of each equation. Place equation in standard form before factoring. Check your solutions and show the check.
9x2 + 4x + 1 = 10x

4) Using the factoring method, solve for the roots of each equation. Place equation in standard form before factoring. Check your solutions and show the check.
(x2/4) - 1 - 2x = - 1

5) Using the factoring method, solve for the roots of each equation. Place equation in standard form before factoring. Check your solutions and show the check.
x2 + 5x = - 2
3

6) Suppose the number of students in a mathematics class is x. The teacher insists that each student participate in group work each class. The number of possible groups is: G = x2 - 3x + 2
2

The class has 17 students. How many possible groups are there?

7) The technology and communication office of a local company has set up a new telephone system so that each employee has a separate telephone and extension number. They are studying the possible number of telephone calls that can be made from people in the office to other people in the office. They have discovered that the total number of possible telephone calls T is described by the equation T = 0.5(x2 - x), where x is the number of people in the office. If 90 people are presently employed at the office, how many possible telephone calls can be made between these 90 people?

7 questions on Quadratic Equations by the square root property:

8) Solve using the square root property; express any complex numbers using i notation: x2 + 25 = 0

9) Solve using the square root property; express any complex numbers using i notation: (7x + 2) 2 = 3

10) Solve the equation by completing the square; completely simplify your answer; express any complex numbers using i notation: x2 - 6x = 13

11) Solve the equation by completing the square; completely simplify your answer; express any complex numbers using i notation: 3y2 - 4y = 4

12) Solve the equation by completing the square; completely simplify your answer; express any complex numbers using i notation:
2x2 + 12x - 7 = - 2

13) Use either the square root property or completing the square. Express any complex numbers using i notation: [(x/3) + 4]2 = 27

14) The time a basketball player spends in the air when shooting a basket is called "the hang time". The vertical leap L measured in feet is related to the hang time t measured in seconds by the equation L = 4t2. A typical athlete has a vertical leap of 1 1/2 to 2 feet; the best male jumpers attain heights of 3 1/2 to 4 feet. What is the hang time of an athlete who has a vertical leap of 3.9 feet?

6 questions on solving a quadratic equations by using the quadratic formula:

15) Solve by the quadratic formula; simplify your answer; use i notation to express any nonreal complex numbers. Clearly label your values for a, b, and c.

9x2 - 12x - 1 = 0

16) Solve by the quadratic formula; simplify your answer; use i notation to express any nonreal complex numbers. Clearly label your values for a, b, and c.

4x2 - 10 = 1

17) Solve by the quadratic formula; simplify your answer; use i notation to express any nonreal complex numbers. Clearly label your values for a, b, and c.

x(x + 3) = (- 4x - 11)/4

18) Solve by the quadratic formula; simplify your answer; use i notation to express any nonreal complex numbers. Clearly label your values for a, b, and c.

1_ + 3 = _2_
12 x x + 1

19) Solve by the quadratic formula; simplify your answer; use i notation to express any nonreal complex numbers. Clearly label your values for a, b, and c.

x2 - 6x + 25 = 0

20) Use the discriminant to find what type of solutions (two rational, two irrational, one rational, or two nonreal complex) the equation has. Do not solve the equation.

4x2 + 9 = 12x

(Full problem set found in attachment).

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20 quadratic equation questions involving word problems and complex roots are solved and discussed in the solution. The solution is detailed and well presented.

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7 questions on Quadratic Equations by Factoring :

1) Using the factoring method, solve for the roots of each equation. Place equation in standard form before factoring. Check your solutions and show the check. 3x2 = 75

Solution:
Divide both sides by 3. ==> x 2 = 25

Subtract 25 on both sides. ==> x 2 - 25 = 0

x 2 - 5 2 = 0 since 5 2 = 25

(x+5)(x-5) = 0 since a 2 - b 2 = (a + b) (a - b)

x = -5 or x = 5 are the solutions.

We then check its solutions by substitution method .

replacing x = 5 3x2 = 3 (5)2 = 3 * 25 = 75 which is the right side.

replacing x = -5 3x2 = 3(-5)2 = 3 * 25 = 75.

2) Using the factoring method, solve for the roots of each equation. Place equation in standard form before factoring. Check your solutions and show the check. 4x2 + 3x = 4x

Solution:
Given our equation as : 4x2 + 3x = 4x

Subtract 4x on both sides. ==> 4x2 + 3x - 4x = 4x - 4x

==> 4x2 - x

Taking x term common outside , x ( 4x -1 ) = 0

x = 0 or (4x-1) = 0
x = 0 or 4x = 1 x = ¼ are its solutions.

check their values using substitution method .

replacing x = 0 4(0) - 3(0) = 4(0)
0 = 0 therefore x = 0 is a solution.

replacing x = ¼ 4 (¼)2 + 3( ¼ ) = 4( ¼ )
4 ( 1/16 ) + ¾ = 1
¼ + ¾ = 1 1 = 1

therefore x = ¼ is also a solution.

3) Using the factoring method, solve for the roots of each equation. Place equation in standard form before factoring. Check your solutions and show the check. 9x2 + 4x + 1 = 10x

Solution:
9x2 + 4x + 1 = 10 x

Subtract 10x on both sides. ==> 9x2 + 4x -10x + 1 = 10x - 10x

9x2 - 6x + 1 = 0

writing -6x as -3x - 3x ==> 9x2 -3x- 3x + 1 = 0

Taking the common terms out : 3x from the first two terms and -1 from the next two,

==> 3x (x-1) - 1 (3x-1) = 0

Then take (x-1) common from both the terms ==> (x - 1) ( 3x - 1) = 0

x - 1 = 0 or (3x-1) = 0

Now resolving for x from them , we get : x = 1 or 3x = 1

==> x = 1 or x = 1/3

we then check for its solution by substitution method

replacing x = 1 9x2 + 4x + 1 = 10x
9(1) + 4(1) + 1 = 10 (1)
9 + 4 + 1 = 10 which is not true.

Therefore x = 1 is not a solution.

Again replacing x = 1/3 9x2 + 4x + 1 = 10x
9 (1/3)2 + 4 (1/3) + 1 = 10 (1/3)
1 + 4/3 + 1 = 10/3
= 10/3 10 /3 = 10 /3

That makes x = 1/3 as it's solution.

4) Using the factoring method, solve for the roots of each equation. Place equation in standard form before factoring. Check your solutions and show the check. (x2/4) - 1 - 2x = - 1

Solution:
Add 1 on both sides. ==> x2/4 - 1 - 2x + 1 = -1 + 1
==> x2/4 - 2x = 0

Taking common terms ' x ' outside , we get : x ( x/4 - 2 ) = 0

Equating them to zero x = 0 or x/4 - 2 = 0
x = 0 or x/4 = 2
x = 0 or x = 8

We then check for its solution, using substitution method .

replacing x = 0 (0) - 1 - 2(0) = -1 -1 = -1

Therefore x = 0 is a ...

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