| solving quadratic equations |
A quadratic equation is an equation in which the variable is to the second degree.
x2 – x – 6 = 0 is a quadratic equation.
The general form of the quadratic equation is ax2 + bx + c = 0 with "a" not equal to zero.
a, b, and c are the coefficients of the variable x.
In other words, the coefficient of the squared term must not be zero or else it is not a quadratic equation.
The coefficients of x to the first and zero powers may be zero as in 6x2 – 3x = 0 and 3x2 – 2 = 0.
A common error is to ignore the signs attached to the coefficients. In the equations above, note that the value of coefficient b is – 3 and not 3. Signs don't float freely in the universe. They're attached to the numbers that follow them.
So, if we have – 5x 2 + 7x – 9 = 0, then a = – 5, b = 7, and c = – 9.
There are three methods of solving quadratic equations.
They are: a) by factoring, b) by completing the square and c) by the quadratic formula.
We generally solve simple quadratics (prime or low value coefficients) by factoring since it is the most efficient method. However, there are quadratics that don't factor easily or at all. In such cases, we can use either completing the square or the quadratic formula to solve for the roots of our quadratic equation. Actually, the quadratic formula is derived from a completion of the square on the general form of the quadratic equation, so both methods are really the same. The only difference is whether you choose to complete the square yourself, or use the template provided by the quadratic formula.
The solutions for a quadratic equation are called its roots. There can be zero, one or two Real number solutions for a quadratic equation.
| intro | factoring | completing the square | quadratic formula | practice | solutions |
The factoring method of solving quadratic equations is based on the axiom that says
| If the product of two or more factors equals zero, then one or more of the factors equals zero. That is: if ab = 0 then a = 0 or b = 0 |
Let's use the factoring method to solve the quadratic equation x2 – x – 6 = 0
| a) We factor the trinomial | (x – 3)(x + 2) = 0 |
| b) Set (x – 3) = 0, solve for x | x – 3 = 0 becomes x = 3 |
| c) Set (x + 2) = 0, solve for x | x + 2 = 0 becomes x = – 2 |
| The solutions are x = 3 and x = – 2 | |
At times, a quadratic equation is presented in a form other than ax2 + bx + c = 0.
It can be ax2 = – bx – c or something else. Since the factoring method for solving quadratic equations is based on setting a product equal to zero, we start by putting the equation into the general form. That is, step one is always relate to zero.
Here's an example:
Solve for x if 3x2 – 10x = 8
| a) Relate to zero | 3x2 – 10x – 8 = 0 |
| b) Factor the trinomial | (3x + 2) (x – 4) = 0 |
| c) Set (3x + 2) = 0 | 3x + 2 = 0 becomes 3x = – 2 becomes x = – 2/3 |
| d) Set (x – 4) = 0 | x – 4 = 0 becomes x = 4 |
| The solutions are x = – 2/3 and x = 4. | |
Any factoring method can be used to solve quadratic equations. When appropriate, we can use common factor, difference of squares, as well as sum or difference of cubes factoring methods to set a product equal to zero.
Here's some examples.
Solve the following quadratic equations
| a) 50x2 = 72 50x2 – 72 = 0 25x2 – 36 = 0 (5x – 6)(5x + 6) = 0 x = 6/5 or x = – 6/5 |
a) Relate to zero. Divide through by 2. Factor the difference of squares. Solve by setting each factor equal to zero. |
| b) 2c2 = – 4c 2c2 + 4c = 0 2c (c + 2) = 0 c = 0 or c = – 2 |
b) Relate to zero. Factor the common factor of 2c. Solve by setting each factor equal to zero. |
.
| intro | factoring | completing the square | quadratic formula | practice | solutions |
.
If we cannot factor our quadratic equation in order to solve it, we can use a technique known as completing the square.
To solve a quadratic equation by completing the square we first express the equation as the square of a binomial equal to a constant, then take the square root of both sides and solve.
Remember that any number has two distinct square roots therefore, when we take square root of both sides, we must express the root of the constant as two values -- the positive and negative roots.
for example: if we have (x – 1) 2 = 25, we will get that (x – 1) = ±5, since when we square them, both – 5 and +5 equal 25.
Let's look at an example and then we'll formulate the technique.
Solve for x if x2 – 6x – 12 = 0
Take ½ of b (– 6), square it, then add it to and subtract it from the left side of the equation to get:
x 2 – 6x + 9 – 9 – 12 = 0
Write the leading trinomial as a perfect square, collect and transpose the constant terms.
(x – 3)2 – 21 = 0 becomes (x – 3)2 = 21
Now we take the square root of both sides. Don't forget that a root can be positive or negative.
x – 3 =
Now we transpose the – 3 to get our two solutions which are:
To solve ax2 + bx + c = 0 by completing the square:
|
note: if you get that the perfect square on the left = a negative value, there are no solutions since every perfect square is positive. (x + 7)² = – 100 has no solutions in the set of Real numbers since the square root of – 100 = 10i which is an imaginary number.
Step 1 is important because we want to write the equation as a perfect square (x – h)2 and set it equal to a constant. Therefore, we need to make the coefficient of x² equal to 1 . If a = 1 as in the previous example, we skip the first step.
Let's do another example in which the coefficient of x2 is not equal to 1.
Solve 2x2 – 3x – 1 = 0 by completing the square.
| Step 1 |  |
| Step 2 |  |
| Step 3 |  |
| Step 4 |  |
| Step 5 |  |
The two solutions are 
Note: in step 2, we took half of 3/2, the coefficient of x, squared it to get 9/16, then added that to and subtracted it from the equation so that we did not change any values. The two sides of the equation remain equal, as they must. We simply change the form of the terms so that we can express the first three terms as a perfect square trinomial which is the square of a binomial.
Really, it's just a slightly more complicated affair than solving x 2 – 4 = 0
We transpose the – 4 to get x 2 = 4 which of course means that x = ± 2
The only tough part is getting the binomial set up -- once there, you're pretty well done.
One more example: Solve 5x2 + 10x – 1 = 0 by completing the square.
| Step 1 divide by 5 |
 |
| Step 2 ½(2) = 1; 1² = 1, so add 1 and subtract 1 |
 |
| Step 3 rewrite as perfect square = constant |
 |
| Step 4 square root both sides |
 |
| Step 5 transpose +1 to become – 1 |
 |
| intro | factoring | completing the square | quadratic formula | practice | solutions |
The quadratic formula is a formula for finding the solutions or zeroes of ax2 + bx + c = 0 with "a" not equal to zero; which is the general form of the quadratic equation. The formula is derived by completing the square on ax2 + bx + c = 0, just as we did in the two previous examples.
| The quadratic formula states that the solutions or zeroes of the quadratic equation ax2 + bx + c = 0 are
|
Let's do the previous example again using the quadratic formula.
Solve by using the quadratic formula: 2x2 – 3x – 1 = 0
In this case, a = 2, b = – 3, and c = – 1.
Substituting these values into the formula we get
The Discriminant
The term b2 – 4ac in the quadratic formula is called the discriminant and it is perfectly named since its value determines the nature of the solutions.
There are three possibilities for the number of solutions for a quadratic equation in the set of Real Numbers. There could be two equal solutions (or really just one solution), two distinct and unequal solutions or no solutions at all. Which of these three possibilities applies in any given quadratic equation depends on the value of the equation's discriminant.
| The quadratic equation ax2 + bx + c = 0 has one solution if b2 – 4ac = 0, two solutions if b2 – 4ac > 0, no real solutions if b2 – 4ac < 0. |
Let's discuss these three possibilities.
1) If b2 – 4ac = 0 then x1 = x2 since 
,
so there is only one solution.
2) If b2 – 4ac > 0 then x1 and x2 are two different values since
3) If b2 – 4ac < 0 then 
is the square root of a negative number and therefore not a Real number. Solutions in the set of Complex or Imaginary numbers exist but in most cases, we are looking for Real number solutions.
Let's look at an example of each of these possible situations
Determine the nature of the solutions of the quadratic equation by finding the discriminant:
a) 9x2 – 12x + 4 = 0
a = 9, b = – 12, c = + 4, so b2 – 4ac = (–12)2 – 4(9)(4) = 144 – 144 = 0
This equation has a single solution which equals 
b) x2 + 5x + 6 = 0
a = 1, b = 5, c = 6, so b2 – 4ac = (5)2 – 4(1)(6) = 25 – 24 = 1
This equation has two solutions which are 
c) x2 + 5x + 8 = 0
a = 1, b = 5, c = 8, so b2 – 4ac = (5)2 – 4(1)(8) = 25 – 32 = – 7
This equation has no real number solutions since the discriminant is a negative number. In the set of Real numbers, we cannot take an even root of a negative value.
When solving word problems with quadratic equations, we must consider the constraints of reality on our solutions. If asked to find the dimensions of a rectangle, an answer of – 2 for the width or length is ridiculous. However, if you can demonstrate a rectangle with a width or length of – 2, the world will hail you as a genius. Until then, let's stick to logical answers.
Had the question been about integers, then – 2 is an acceptable answer since – 2 is an integer.
| intro | factoring | completing the square | quadratic formula | practice | solutions |
A/ Solve these quadratic equations by factoring:
1) x2 + 4x = 21
2) 8y2 + 2y = 1
B/ Solve these quadratic equations by completing the square:
1) x 2 – 6x = – 3
2) y 2 – 2y = 2
3) 2x 2 = 3 – x
C/ Use the quadratic formula to solve these equations:
1) x 2 + 5x = 3
2) 9x 2 – 6x + 1 = 0
3) 3x 2 – 2x = – 1
D/ Word Problems solved with quadratics:
1) Find 3 consecutive integers such that the square of the 2nd increased by the product of the 1st and the 3rd, is 199.
2) The base of a triangle is 6 cm. longer than its altitude or height. Its area is 80 square centimetres. What are its dimensions?
3) The perimeter of a rectangular field is 50 metres.
What are its dimensions if its area is 150 square metres?
| intro | factoring | completing the square | quadratic formula | practice | solutions |
A/ Solve these quadratic equations by factoring:
| 1) x 2 + 4x = 21 x 2 + 4x – 21 = 0 (x + 7)(x – 3) = 0 x = –7 or x = 3 |
2) 8y 2 + 2y = 1 8y 2 + 2y – 1 = 0 (4y – 1)(2y + 1) = 0 y = 1/4 or y = – 1/2 |
B/ Solve these quadratic equations by completing the square:
| 1) x 2 – 6x = – 3 x 2 – 6x + 3 = 0 x 2 – 6x + 9 – 9 + 3 = 0 (x – 3) 2 = 6 x – 3 = ! x = 3 ! |
2) y 2 – 2y = 2 y 2 – 2y – 2 = 0 y 2 – 2y + 1 – 1 – 2 = 0 (y – 1) 2 = 3 y – 1 = ! y = 1 ! |
3) 2x 2 = 3 – x 2x 2 + x – 3 = 0 2(x 2 + ½x + 2(x + (x + x + or x = |
C/ Use the quadratic formula to solve these equations:
1) x 2 + 5x = 3
there are 2 real roots |
2) 9x 2 – 6x + 1 = 0
there is a unique real root |
| 3) 3x2 – 2x = – 1 3x2 – 2x + 1 = 0 a = 3, b = – 2, c = 1 |
D/ Word Problems solved with quadratics:
1) Let x = the 1st integer
2)
Let h = the height of the triangle
therefore h + 6 = the base
The area of a triangle is found by multiplying base % height + 2. Since area = 80 cm 2 :
so, h (h + 6) = 160 becomes h 2 + 6h – 160 = 0
(h + 16) (h – 10 ) = 0
so, h = – 16 or h = 10
If you can build a triangle with a height equal to –16 cm -- the world will applaud you.
Until then, the answer to this question will be that the height is 10 cm. and the base is 16 cm.
.
3) The perimeter of a rectangular field is 50 metres.
What are its dimensions if its area is 150 square metres?
If the perimeter, the sum of 4 sides, is equal to 50 m. then 2 sides must add to 25 m.
Let x = the width
therefore, 25 – x = the length.
We know that the area is 150 m 2 so that's what our equation must state:
| intro | factoring | completing the square | quadratic formula | practice | solutions |
.
(all content © MathRoom Learning Service; 2004 - ).
1, multiply through by 1/a, to obtain the form x2 + (b/a)x + c/a = 0. 
.
) – 3 = 0
)2 =
) = 
becomes x = 1
there are no real roots