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Year & Section: _________________Teacher: _______________

Reviewer: Quadratic Equations

I. Multiple Choice: Choose the letter of the correct answer. Show your solution.

1. What are the values of x that satisfy the equation 3 – 27x2 = 0? A. x = [pic]3B. x = [pic]C. x = [pic]D. x = [pic] 2. What are the solutions of the equation 6x2 + 9x – 15 = 0? A. 1, - 15B. 1, [pic]C. – 1, - 5D. 3, [pic]

3. For which equation is – 3 NOT a solution?
A. x2 – 2x – 15 = 0C. 2x2 + 12x = - 18
B. x2 – 21 = 4xD. 9 + x2 = 0

4. What are the solutions of the equation (2x – 7)2 = 25? A. 6, - 6B. 6, 1C. 6, -1D. – 6, - 1

5. Find the sum of the solutions to the equation x2 + 2x – 15 = 0. A. 8B. – 8C. 2D. – 2

6. Find the product of the solutions to the equation x2 – 8x = 9. A. 6B. – 6C. 9D. – 9
7. Which equation has [pic]as a solution?
A. (2x – 5)(x + 1) = 0C. (5x + 2)(x + 1) = 0
B. 5x – 2)(x + 1)D. (2x + 5)(x + 1)

8. The equation x2 – 3x + a = 0 has two roots. One root of the equation is 2. What is the other root? A. – 2B. – 1C. 1D. 3

9. What is the quadratic equation determine by the roots 3 and – 4 ? A. x2 + x – 12 = 0C. x2 + x + 12 = 0
B. x2 – x – 12 = 0D. x2 – x + 12 = 0

10. One solution to the equation x2 + bx – 20 = 0 is 5. What is the other solution? A. 4B. – 4C. [pic]D. [pic]
11. What can you add to x2 + 5x to get a perfect square trinomial? A. [pic]B. [pic]C. 2.5D. 25
12. How can you rewrite the equation x2 + 12x + 5 = 3 so the left side of the equation is in the form (x + a)2? A. (x – 6)2 = 28B. (x + 6)2 = 39C. (x + 6)2 = 34D. ( x + 12)2 = - 2

13. How many real solutions does x2 – 10x + 25 = 0 have?? A. No solutionsB. One solutionC. Two solutionsD. Many...

...as prioritization of traffic, maximizing bandwidth utilization or to prevent network congestions and bottlenecks. LSP routes are most popularly used to create Virtual Private Networks (VPN).
The main thing to keep in mind about MPLS is that it’s a networking technique and not a service. Being that MPLS is a packet switching networking technique that makes its entire packet forwarding decisions based on the label and removing the need to examine the entire packet, this allows the creation of end-to-end circuits across any medium using any protocol. This process eliminates the dependence on specific data link layer technologies like Asynchronous Transfer Mode (ATM), Frame Relay, Synchronous Optical Networking (SONET) or Ethernet. Furthermore, this eliminates the need for multiple layer 2 networks for the different types of traffic like Data, Voice and Video and allowing you to combine all into a single network.
QoS over MPLS
Now that we understand how the MPLS networking techniques work, we can now focuses on the QoS capabilities it provides. However, before moving forward we need to understand why QoS is needed in an IP network. IP networks are not connection oriented in nature and are considered best effort delivery technology, it does not care about traffic priorities or bandwidth requirements it treats all traffic the same across a pure IP network.
It is important to understand that MPLS does not define a new QoS technique. However,...

...329
QuadraticEquations
Chapter-15
QuadraticEquations
Important Definitions and Related Concepts
1. QuadraticEquation
If p(x) is a quadratic polynomial, then p(x) = 0 is called
a quadraticequation. The general formula of a quadraticequation is ax 2 + bx + c = 0; where a, b, c are real
numbers and a 0. For example, x2 – 6x + 4 = 0 is a
quadraticequation.
2. Roots of a QuadraticEquation
Let p(x) = 0 be a quadraticequation, then the values of
x satisfying p(x) = 0 are called its roots or zeros.
For example, 25x2 – 30x + 9 = 0 is a quadraticequation.
3
And the value of x =
is the solution of the given
5
equation.
3
Since, if we put x =
in 25x2 – 30x + 9 = 0, we have,
5
2
3
3
LHS = 25 × – 30 ×
+ 9
5
5
= 9 – 18 + 9 = 0 = RHS
Finding the roots of a quadraticequation is known as
solving the quadraticequation.
5. Methods of Solving QuadraticEquation
( i ) By Factorization
This can be understood by the examples given
below:
2
Ex. 1: Solve: 25 x 30 x 9 0
Soln: 25x 2 30x 9 0 is equivalent to
5x 2 25x 3 32
0
5 x 32 0
3 3
3
,
or simply x
as the
5 5
5
required...

...QUADRATICEQUATIONSQuadraticequations Any equation of the form ax2 + bx + c=0, where a,b,c are real numbers, a 0 is a quadraticequation.
For example, 2x2 -3x+1=0 is quadraticequation in variable x.
SOLVING A QUADRATICEQUATION
1.Factorisation
A real number a is said to be a root of the quadraticequation ax2 + bx + c=0, if aa2+ba+c=0. If we can factorise ax2 + bx + c=0, a 0, into a product of linear factors, then the roots of the quadraticequation ax2 + bx + c=0 can be found by equating each factor to zero.
Example – Find the roots of the equation 2x2 -5x +3=0, by factorisation.
Solution:
2x2 -5x +3=0 2x2 -2x-3x+3=0 2x(x-1)-3(x-1)=0 i.e., (2x-3)(x-1)=0 Either 2x-3=0 or x-1=0. So,the roots of the given equation are x=3/2 and x=1.
2. Completing the square
To complete the square means to convert a quadratic to its standard form. We want to convert ax2+bx+c = 0 to a statement of the form a(x h)2 + k = 0.
To do this, we would perform the following steps:
1) Group together the ax2 and bx terms in parentheses and factor out the coefficient a.
2) In the parentheses, add and subtract (b/2a)2, which is half of the x coefficient, squared.
3) Remove the term - (b/2a)2 from the parentheses....

...QuadraticEquation:
Quadraticequations have many applications in the arts and sciences, business, economics, medicine and engineering. QuadraticEquation is a second-order polynomial equation in a single variable x.
A general quadraticequation is:
ax2 + bx + c = 0,
Where,
x is an unknown variable
a, b, and c are constants (Not equal to zero)
Special Forms:
* x² = n if n < 0, then x has no real value
* x² = n if n > 0, then x = ± n
* ax² + bx = 0 x = 0, x = -b/a
WAYS TO SOLVE QUADRATICEQUATION
The ways through which quadraticequation can be solved are:
* Factorizing
* Completing the square
* Derivation of the quadratic formula
* Graphing for real roots
Quadratic Formula:
Completing the square can be used to derive a general formula for solving quadraticequations, the quadratic formula. The quadratic formula is in these two forms separately:
Steps to derive the quadratic formula:
All QuadraticEquations have the general form, aX² + bX + c = 0
The steps to derive quadratic formula are as follows:
Quadraticequations and functions are very important in business...

...Quadraticequation
In elementary algebra, a quadraticequation (from the Latin quadratus for "square") is any equation having the form
where x represents an unknown, and a, b, and c represent known numbers such that a is not equal to 0. If a = 0, then the equation is linear, not quadratic. The numbers a, b, and c are the coefficients of theequation, and may be distinguished by calling them, the quadratic coefficient, the linear coefficient and the constant or free term.
Solving the quadraticequation
A quadraticequation with real or complex coefficients has two solutions, called roots. These two solutions may or may not be distinct, and they may or may not be real.
Factoring by inspection
It may be possible to express a quadraticequation ax2 + bx + c = 0 as a product (px + q)(rx + s) = 0. In some cases, it is possible, by simple inspection, to determine values of p, q, r, and s that make the two forms equivalent to one another. If the quadraticequation is written in the second form, then the "Zero Factor Property" states that the quadraticequation is satisfied if px + q = 0 or rx + s = 0. Solving these two linear equations provides the roots...

...While the ultimate goal is the same, to determine the value(s) that hold true for the equation, solving quadraticequations requires much more than simply isolating the variable, as is required in solving linear equations. This piece will outline the different types of quadraticequations, strategies for solving each type, as well as other methods of solutions such as Completing the Square and using theQuadratic Formula. Knowledge of factoring perfect square trinomials and simplifying radical expression are needed for this piece. Let’s take a look!
Standard Form of a QuadraticEquation
ax2+ bx+c=0
Where a, b, and c are integers
and a≥1
I. To solve an equation in the form ax2+c=k, for some value k. This is the simplest quadraticequation to solve, because the middle term is missing.
Strategy: To isolate the square term and then take the square root of both sides.
Ex. 1) Isolate the square term, divide both sides by 2
Take the square root of both sides
2x2=40
2x22= 40 2
x2 =20
Remember there are two possible solutions
x2= 20
Simplify radical; Solutions
x= ± 20
x=± 25
(Please refer to previous instructional materials Simplifying Radical Expressions )
II. To solve a quadratic...

...QuadraticEquationsEquationsQuadratic
MODULE - I
Algebra
2
Notes
QUADRATICEQUATIONS
Recall that an algebraic equation of the second degree is written in general form as
ax 2 + bx + c = 0, a ≠ 0
It is called a quadraticequation in x. The coefficient ‘a’ is the first or leading coefficient, ‘b’
is the second or middle coefficient and ‘c’ is the constant term (or third coefficient).
For example, 7x² + 2x + 5 = 0,
5
1
x² + x + 1 = 0,
2
2
1
= 0, 2 x² + 7x = 0, are all quadraticequations.
2
In this lesson we will discuss how to solve quadraticequations with real and complex
coefficients and establish relation between roots and coefficients. We will also find cube
roots of unity and use these in solving problems.
3x² − x = 0, x² +
OBJECTIVES
After studying this lesson, you will be able to:
• solve a quadraticequation with real coefficients by factorization and by using quadratic
formula;
• find relationship between roots and coefficients;
• form a quadraticequation when roots are given; and
• find cube roots of unity.
EXPECTED BACKGROUND KNOWLEDGE
• Real numbers
• QuadraticEquations with real coefficients.
MATHEMATICS
39...

...Section 2.1
Linear and Quadratic Functions and Modeling
67
Chapter 2 Polynomial, Power, and Rational Functions
■ Section 2.1 Linear and Quadratic Functions and Modeling
Exploration 1
1. –$2000 per year 2. The equation will have the form v(t)=mt+b. The value of the building after 0 year is v(0)=m(0)+b=b=50,000. The slope m is the rate of change, which is –2000 (dollars per year). So an equation for the value of the building (in dollars) as a function of the time (in years) is v(t)=–2000t+50,000. 3. v(0)=50,000 and v(16)=–2000(16)+50,000=18,000 dollars 4. The equation v(t)=39,000 becomes –2000t+50,000=39,000 –2000t=–11,000 t=5.5 years
6. (x-4)2=(x-4)(x-4)=x2-4x-4x+16 =x2-8x+16 7. 3(x-6)2=3(x-6)(x-6)=(3x-18)(x-6) =3x2-18x-18x+108=3x2-36x+108 8. –3(x+7)2=–3(x+7)(x+7) =(–3x-21)(x+7)=–3x2-21x-21x-147 =–3x2-42x-147 9. 2x2-4x+2=2(x2-2x+1)=2(x-1)(x-1) =2(x-1)2 10. 3x2+12x+12=3(x2+4x+4)=3(x+2)(x+2) =3(x+2)2
Section 2.1 Exercises
1. Not a polynomial function because of the exponent –5 2. Polynomial of degree 1 with leading coefficient 2 3. Polynomial of degree 5 with leading coefficient 2 4. Polynomial of degree 0 with leading coefficient 13 5. Not a polynomial function because of cube root 6. Polynomial of degree 2 with leading coefficient –5 5 5 5 18 7. m= so y-4= (x-2) ⇒ f(x)= x+ 7 7 7 7
y 5 (2, 4) x
Quick Review 2.1
1. y=8x+3.6 2. y=–1.8x-2 3 3. y-4=– (x+2), or y=–0.6x+2.8 5
y 7
3...