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A L G E B R A

37

The standard form of a quadratic equation

Section 2:

Completing the square

Section 3:

The graph of  y = A quadratic:  A parabola

A QUADRATIC is a polynomial whose highest exponent is 2.

ax² + bx + c.

The coefficient of x² is called the leading coeffieient.

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

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, the roots of this quadratic --

x² + 2x − 8

-- are the solutions to

x² + 2x − 8 = 0.

To find the roots, we can factor that quadratic as

(x + 4)(x − 2).

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

x² + 2x − 8 = 0.

−4  or  2 are the solutions to the quadratic equation. They are the roots of that quadratic.

Conversely, if the roots are a or b say, then the quadratic can be factored as

(xa)(xb).

A root of a quadratic is also called a zero. Because, as we will see, at each root the value of the graph is 0. (See Topic 7 of Precalculus, Question 2.)

Question 3.  How many roots has a quadratic?

Always two. Because a quadratic (with leading coefficient 1, at least) can always be factored as (xa)(xb), and a, b are the two roots.

In other words, when the leading coefficient is 1, the root has the opposite sign of the number in the factor.
If (x + q) is a factor, then  x = −q  is a root.

q + q = 0.

Problem 1.   If a quadratic can be factored as (x + 3)(x − 1), then what are the two roots?

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−3 or 1.

We say "or," because x can take only one value at a time.

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

The two roots are equal. The factors will be (xa)(xa), so that the two roots are a, a.

For example, this quadratic

x² − 12x + 36

can be factored as

(x − 6)(x − 6).

If x = 6, then each factor will be 0, and therefore the quadratic will be 0.  6 is called a double root.

When will a quadratic have a double root?  When the quadratic is a perfect square trinomial.

Example 1.   Solve for x:   2x² + 9x − 5.

Solution.   That quadratic is factored as follows:

2x² + 9x − 5 = (2x − 1)(x + 5).

Now, it is easy to see that the second factor will be 0 when x = −5.

As for the value of x that will make

 2x − 1 = 0, we must solve that little equation. (Lesson 9.) We have: 2x = 1 x = 12

The solutions are:

 x = 12 or  −5.

Problem 2.   How is it possible that the product of two factors ab = 0?

Either a = 0 or b = 0.

Solution by factoring

Problem 3.   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.

Again, 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 = − 12 or  −3. x = 23 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 2.   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 = − ba .

Those are the two roots.

Problem 4.   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  − 43 x = 0  or  ½

Example 3.   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 = ca x = ,  according to Lesson 26.

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

x² − 16

-- then we can factor it as:

(x + 4)(x −4).

The roots are ±4.

In fact, if the quadratic is

x² − c,

then we could factor it as:

(x + )(x ),

so that the roots are  ± .

Problem 5.   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 4.   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.

And so an equation is solved when x is isolated on the left.

x = ± is not a solution.

Problem 6.   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 5.   Solve this equation

 3 − 52 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 + 52 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 23 or − 32 .

Problem 7.   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 25 or − 32 . The roots are 43 or − 35 .
 c) −x² − x + 20 = 0 d) −x² + 3x + 18 = 0 x² + x − 20 = 0 x² − 3x − 18 = 0 (x + 5)(x − 4) = 0. (x − 6)(x + 3) = 0. x = −5  or  4. x = 6  or  −3.

Section 2:  Completing the square Please make a donation to keep TheMathPage online.
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