How do quadratic equations work?

Answer 1

quadratic equations don't work. they are on unemployment.

Answer 2

S k i l l
i n
A L G E B R A

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37

QUADRATIC EQUATIONS

Solution by factoring

Section 2

Completing the square

The quadratic formula

The discriminant

Proof of the quadratic formula

Section 3

The graph of y = A quadratic



A QUADRATIC is a polynomial whose highest exponent is 2.

Question 1. What is the standard form of a quadratic equation?

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ax² + bx + c = 0

The quadratic is on the left. 0 is on the right.



Question 2. What do we mean by a root of a quadratic?

A solution to the quadratic equation.

For example, this quadratic

x² + 2x − 8

can be factored as

(x + 4)(x − 2).

Now, if x = −4, then the first factor will be 0. While if x = 2, the second factor will be 0. But if any factor is 0, then the entire product will be 0. That is, if x = −4 or 2, then

x² + 2x − 8 = 0.

Therefore, −4 and 2 are the roots of that quadratic. They are the solutions to the quadratic equation.

A root of a quadratic is also called a zero. Because, as we will see, at those values of x, the graph has the value 0.

Question 3. How many roots has a quadratic?

Always two.

Question 4. What do we mean by a double root?

The two roots are equal.

For example, this quadratic

x² − 10x + 25

can be factored as

(x − 5)(x − 5).

If x = 5, then each factor will be 0, and therefore the quadratic will be 0. 5 is called a double root.

A quadratic will have a double root if the quadratic is a perfect square trinomial.

Problem 1. If either a = 0 or b = 0, then what can you conclude about ab ?

ab = 0

Solution by factoring

Problem 2. Find the roots of each quadratic by factoring.

a) x² − 3x + 2 b) x² + 7x + 12

(x − 1)(x − 2) (x + 3)(x + 4)

x = 1 or 2. x = −3 or −4.

Notice that we use the conjunction "or," because x takes on only one value at a time.

c) x² + 3x − 10 d) x² − x − 30

(x + 5)(x − 2) (x + 5)(x − 6)

x = −5 or 2. x = −5 or 6.
e) 2x² + 7x + 3 f) 3x² + x − 2

(2x + 1)(x + 3) (3x − 2)(x + 1)

x = − 1
2 or −3. x = 2
3 or −1.
g) x² + 12x + 36 h) x² − 2x + 1

(x + 6)² (x − 1)²

x = −6, −6. x = 1, 1.

A double root. A double root.

Example 1. c = 0. Solve this quadratic equation:

ax² + bx = 0

Solution. Since there is no constant term -- c = 0 -- x is a common factor:

x(ax + b) = 0.

This implies:

x = 0

or

x = − b
a .

Those are the two roots.

Problem 3. Find the roots of each quadratic.

a) x² − 5x b) x² + x

x(x − 5) x(x + 1)

x = 0 or 5. x = 0 or −1.
c) 3x² + 4x d) 2x² − x

x(3x + 4) x(2x − 1)

x = 0 or − 4
3 x = 0 or ½

Example 2. b = 0. Solve this quadratic equation:

ax² − c = 0.

Solution. In the case where there is no middle term, we can write:

ax² = c.

This implies:
x² = c
a

x = , according to Lesson 26.

However, if the form is the difference of two squares --

x² − 16

-- then we can factor:

(x + 4)(x −4)

The roots are ±4.

In fact, if the quadratic is

x² − c,

then we could factor:

(x + )(x − )

so that the roots are ±.

Problem 4. Find the roots of each quadratic.

a) x² − 3 b) x² − 25 c) x² − 10

x² = 3 (x + 5)(x − 5) (x + )(x − )

x = ±. x = ±5. x = ±.

Example 3. Solve this quadratic equation:

x² = x + 20

Solution. First, rewrite the equation in the standard form, by transposing all the terms to the left:

x² − x − 20 = 0

(x + 4)(x − 5) = 0

x = −4 or 5.

Thus, an equation is solved when x is isolated on the left.
x = ± is not a solution.

Problem 5. Solve each equation for x.

a) x² = 5x − 6 b) x² + 12 = 8x

x² − 5x + 6 = 0 x² − 8x + 12 = 0

(x − 2)(x − 3) = 0 (x − 2)(x − 6) = 0

x = 2 or 3. x = 2 or 6.

c) 3x² + x = 10 d) 2x² = x

3x² + x − 10 = 0 2x² − x = 0

(3x − 5)(x + 2) = 0 x(2x − 1) = 0

x = 5/3 or − 2. x = 0 or 1/2.

Example 4. Solve this equation

3 − 5
2 x − 3x² = 0

Solution. We can put this equation in the standard form by changing all the signs on both sides. 0 will not change. We have the standard form:

3x² + 5
2 x − 3 = 0

Next, we can get rid of the fraction by multiplying both sides by 2. Again, 0 will not change.

6x² + 5x − 6 = 0

(3x − 2)(2x + 3) = 0
The roots are 2
3 and − 3
2 .

Problem 6. Solve for x.

a) 3 − 11
2 x − 5x² = 0 b) 4 + 11
3 x − 5x² = 0

5x² + 11
2 x − 3 = 0 5x² − 11
3 x − 4 = 0

10x² + 11x − 6 = 0 15x² − 11x − 12 = 0

(5x − 2 )(2x + 3) = 0 (3x − 4)(5x + 3 ) = 0


The roots are 2
5 and − 3
2 . The roots are 4
3 and − 3
5 .

Section 2



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Answer 3

A quadratic equation is one in which the variable is raised to the second power. Examples are y= x^2 +x +1, y= x^2, x= y^2, A=s^2, V=-16t^2+8t +5

The general equation is ax^2 +bx + c =0, where a,b, and c are constants and a not = 0.

A quadratic equation always has exactly two solutions. They may be two different real numbers, two different complex or pure imaginary numbers, or the same number repeated twice.

If you graph a quadratic equation, you get a curve called a parabola. You should study parabolas in your text book or on the internet as they are used frequently in real life and mathematics.

Answer 4

The quadratic equation works because it comes from solving
ax^2 + bx + c = 0
where a, b, and c represent any real number.
So if you work that out and solve for x using the variables a, b, and c, you will get the quadratic formula:
x = [ -b ± √(b^2-4ac) ] / 2a

There are formulas for polynomials with degree 3 and 4, but are more complicated. Nobody has discobered the formula for a polynomial with degree 5 or greater because of the extreme complexity.

Answer 5

Quatratic equations can work by simply factorizing and solving for the variable or by using the quadratic formula.

Answer 6

Quadratic equations are called quadratic because quadratus is Latin for "square"; in the leading term the variable is squared.