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37

QUADRATIC EQUATIONS

Solution by factoring

Section 2

Completing the square

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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