Objectives:  To recall how to solve equations like 3x^2 - 5x + 2 = 0.

Recall 1:  A quadratic equation is an equation of the form

ax^2 + bx + c = 0,

where a, b and c are fixed rational numbers and a is not 0.

Recall 2:  Many quadratic equations may factor as

(x - r1)(x - r2)= 0,

where r1 and r2 are rational numbers.   By rewriting a   quadratic equation in this form, we find the solution to be  x = r1 and x = r2.

For example, the equation

x^2 - 3x + 2 = 0

may be rewritten as

(x - 1)(x - 2) = 0.

This means that x = 1 and x = 2 are the solutions to x^2 - 3x + 2 = 0.

Other quadratic equations may not split up as
(x - r1)*(x - r2)= 0,
unless we relax the restriction that r1  and r2 must be rational numbers.  For such equations we   may have to be satisfied with the possibility that r1 and r2   may be real, and not rational.

For example x^2 - x - 1 = 0 cannot be rewritten as

(x - r1)(x - r2)= 0,

with r1 and r2 being rational.

The factors of   x^2 - x - 1 may be written as

But neither (1 + sqrt(5)) / 2, nor (1 - sqrt(5)) / 2 is a rational number.

Still, other quadratic equations may not split up as
(x - r1)(x - r2)= 0,
even if we allow r1 and r2 to be real   numbers.
For example, x^2 + x + 1 = 0 cannot be written as   (x - r1)(x - r2) for any real numbers r1 and r2.

Recall 3:  Even though some quadratic equations may be split into a   product of linear terms as (x - r1)(x - r2), it may not always be obvious what the values of r1 and r2 should be.
Here is a way to find the values of r1 and r2.

In splitting the quadratic x^2 - 3x + 2 = 0, we are looking  for an r1 and r2 that fit into

x^2 - 3x + 2 = (x - r1)(x - r2).

By multiplying out the right-hand-side, we find that

x^2 - 3x + 2 = x^2 - (r1 + r2)x + r1*r2

By comparing the right side with the left side, we find that

3 = r1 + r2 and
2 = r1*r2

If we want r1 to be an integer, then it must be ±2 or ±1, for the only two integers that multiply together to give 2 are 2   and 1.  Test ±2 and ±1 to find the right combination that   gives r1 + r2 = 3 and r1*r2 = 2. You can see that r1 = 1 and   r2 = 2 is the right choice.  Hence, we know that

x^2 - 3x + 2 = (x - 1)(x - 2)

and therefore that x = 1 and x = 2 are solutions to the   equation x^2 - 3x + 2 = 0.

Note

It is important to check your solutions by substituting them into the original equations.  In the case of the solutions x = 1 and x = 2 above, we check that
(1)^2 - 3(1) + 2 = 0. and that (2)^2 - 3*(2) + 2 = 0.
In general, checking is easy and worth the little extra time.

Recall 4:  If the factors of a quadratic equation are not immediately   apparent, you may use the quadratic formula to find the solutions to

ax^2 + bx + c = 0.

For example, the equation x^2 - x - 1 = 0 has solutions

This means that there are two solutions to the equation

x^2 - x - 1 = 0,

namely,   x= (1 + sqrt(5)) / 2  and  x= (1 - sqrt(5)) / 2.

Examples

Example 1. Find the solutions to the equation

x^2 + 5x + 6 = 0.

Solution:   Look for two integers whose product is 6 and whose sum is -5.  The factors of 6 are

±2 and ±3
±1 and ±6;

so, candidates for r1 and r2 are

r1 = 2 and r2 = 3.
r1 = -2 and r2 = -3.
or

r1 = 1 and r2 = 6.
r1 = -1 and r2 = -6.

Next, we test the product

(x + 2)(x + 3),

to see if it is equal to x^2 + 5x + 6.   It is.  Therefore, x = -2   and x = -3 are the solutions to the equation in question.

Example 2. Find the solutions to the equation

2x^2 + 10x + 12 = 0.

Solution:    After factoring out the 2, the equation becomes

2(x^2 + 5x + 6) = 0

Divide both sides by 2 to get the equivalent equation

x^2 + 5x + 6 = 0.

This is the equation from Example 1 and, therefore, it   has the same solutions x = -2 and x = -3.

Example 3. Find the solutions to the equation

3x^2 + 7x - 6 = 0.

Solution:    This example involves a case that is different than any that we have investigated so far.  The leading term is a 3, not a 1.  This poses a small difficulty, not an overwhelming one.  We begin by noticing that the only way to get a 3 as the leading term is if one of the factors of 3x^2 + 7x - 6 has a factor of 3 as its leading term.  The only factors of 3 are 3 and 1; so, we look for factors of the form

(3x - r1)(x - r2).

The factors of 6 are

±2 and ±3
±1 and ±6;

so, candidates for r1 and r2 are

r1 = 2 and r2 = -3.
r1 = -2 and r2 = 3.
or

r1 = 1 and r2 = -6.
r1 = -1 and r2 = 6.

Next, we test the product with one of our candidates - say r1 = -1 and r2 = 6:

(3x - (-1))(x - 6),

to see if it is equal to 3x^2 + 7x - 6.   It is not.  Therefore, we must test  another possible set of values for r1 and r2 , say r1 = 2 and r2 = -3.  This  means that our factors have the form

(3x - 2)(x - (-3)),
or
(3x - 2)(x + 3).

Test this last product to see that it multiplies out to

3x^2 + 7x - 6

This means that 3x - 2 = 0 and x + 3 = 0.  We may conclude that the solutions are

x =2/3  and x = -3.

Note

In Example 3, we used the following fact:

If a product of terms equals zero, then one or both of the terms must be zero.

We used that fact to conclude that if (3x - 2)(x + 3) = 0, then   3x - 2 = 0 or x + 3 = 0.  This is how we concluded  that x = 2/3  or x = -3.

Example 4.  Find the solutions to the equation

2x^2 + 3x - 3 = 0.

Solution:      If you search for factors of the form

(2x - r1)(x - r2)

where r1 and r2 are rational numbers, you will not find them.  There are none.  In this case, the best you could do is use the quadratic formula by comparing 2x^2 + 3x - 3 = 0 with ax^2 + bx + c = 0.  Notice that these last two quadratic equations are the same when a = 2, b = 3 and c = -3. (Note that  c must be -3, not +3.)  Then, the solutions are

where a = 2, b = 3 and c = -3.  Thus, x reduces to

Notice that these solutions are not rational.

Note

How do you know when to look for factors of the quadratic and when to use the quadratic formula?  You may always use the quadratic formula.  Even in the cases where the equation factors.  However, to check if the formula is necessary, try computing the quantity b^2 - 4ac.  If this quantity is a perfect square, then the factors will be rational numbers; otherwise, the factors will not be rational.  The next example shows what happens if you use the quadratic formula on an equation that has simple integer factors.

Example 5.  Find the solutions to the equation

2x^2 + 3x - 2 = 0.

Solution:     If you compute the value of b^2 - 4ac, as suggested above, you find that it is equal to 25, since b = 3, a = 2 and c = -2.  This tells us that b^2 - 4ac is a perfect square, since 25 = 5^2.  So the quadratic equation factors.  Indeed, it factors as

(2x - 1)(x + 2) = 0,

from which we may conclude that x =  and x = -2.

But, what if we used the quadratic formula

to find the solutions to 2x^2 + 3x - 2 = 0 ?

In this case a = 2, b = 3 and c = -2.  Thus, x reduces to

So, we conclude that x = 1/2 or x = -2.  Notice that these are the same solutions that we computed from the factoring of the equation.    This means that you are always welcome to use the quadratic formula, if all else fails.

EXERCISES

Solve the following quadratic equations for the appropriate variable:

1. x^2 - x - 2 = 0

2. x^2 - 7x + 12 = 0

3  x^2 + 2x - 8 = 0

4 2x^2 + 6x + 4 = 0

5 2x^2 - x - 1 = 0

6 3x^2 + 7x + 2 = 0

7.  x^2 + 3x - 5 = x + 3  Hint:  Subtract (x + 3) from both sides.

8.  3x^2 - 1 = x^2 + x              Hint:  Collect all terms to the left side.

9.  x^2 + x - 1 = 0

10. 3x^2 - 2x - 4 = 0

11.  5x^2 = 5 - 2x

12.  3x^2 + 4x + 1 = 0

13.  6x^2 - 5x - 6 = 0

14.  15x^2 - 16x - 15 = 0

15.  x^2 + 0.9x - 2.52 = 0

16.  25x^2 -  4 = 0

17. 4x^2 - 20x + 25 = 0