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37

QUADRATIC EQUATIONS

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  and  2 are the solutions to the quadratic equation. They are the roots of that quadratic.

Conversely, if the roots are a and 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).

Lesson 17.

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 = 1
2

The solutions are:

x = 1
2
 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 = − 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 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   =   b
a
.

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  − 4
3
  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 = c
a
 
  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 −  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 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  2
5
 and − 3
2
.   The roots are  4
3
 and − 3
5
.
   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.   x6  or  −3.

Section 2:  Completing the square

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