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Quadratic Functions and Equations

Solving Quadratic Equations

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Concept

Quadratic Equation

A quadratic equation is a polynomial equation of second degree, meaning the highest exponent of its monomials is 2. A quadratic equation can be written in standard form as follows.

ax2+bx+c=0

Here, the leading coefficient a is non-zero, which guarantees that the x2-term is present. Solving a quadratic equation means finding the zeros of the related quadratic function. Therefore, quadratic equations can have at most two solutions.
This type of equation can be solved using several methods, such as graphing and factoring.

Method

Solving a Quadratic Equation Graphically

A quadratic equation can be solved graphically by drawing the parabola that corresponds to the related quadratic function and identifying its zeros. Consider the following equation as an example.
These four steps can be followed to find the solutions of the equation.
1
Write the Equation in Standard Form
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If neither side of the equation equals 0, rearrange it so that all terms are on the same side.
If there is more than one term with the same degree, simplify the equation by combining like terms.
The equation is in standard form now.
2
Graph the Related Function
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The side with the variables can now be seen as a function f(x).
To graph the quadratic function in standard form, its characteristics should be identified first.
f(x)=2x22x12
Direction Vertex Axis of Symmetry y-intercept
a>0 upward (0,-12)

In addition to these points, the reflection of the y-intercept across the axis of symmetry is at (1,-12), which is also on the parabola. Now, the graph can be drawn.

Graph of related function
3
Identify Any Points With y-coordinate 0
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Now, find all the points on the graph that have a y-coordinate equal to 0.

4
Identify the x-coordinates of Those Points
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The x-coordinates of the identified points solve the equation f(x)=0.

In the example, the x-coordinates of the points where the curve intercepts the x-axis are -2 and 3. Note that solving an equation graphically does not necessarily lead to an exact answer. To verify a solution, substitute it into f(x) and evaluate the expression. Verifying the solution x=-2 is done by evaluating f(-2).
f(x)=2x22x12
f(-2)=2(-2)22(-2)12
Evaluate right-hand side
f(-2)=2(4)2(-2)12
f(-2)=82(-2)12
f(-2)=8(-4)12
f(-2)=8+412
f(-2)=0
When x=-2, the value of f(x) is 0. Therefore, x=-2 is an exact solution. The solution x=3 can now be verified by following the same procedure.
Solution Substitute Evaluate
x=-2 f(-2)=2(-2)22(-2)12
x=3 f(3)=2(3)22(3)12
Since f(3)=0, x=3 is also a solution. Therefore, the equation 2x2+3x5=5x+7 has two solutions.

Theory

Simple Quadratic Equation

Simple quadratic equations take the form
ax2+c=0
and are solved using inverse operations. Once x2 remains on the left-hand side, the equation can be written as x2=d, where . The value of d gives the number of solutions the equation has.

Method

Solving Quadratic Equations with Square Roots

Simple quadratic equations are quadratic equations whose linear coefficient b is equal to 0.
This type of equation can be solved using inverse operations, and two steps must be followed.
  1. Isolate x2.
  2. Take square roots on both sides of the equation.
As an example, consider the equation 5x2500=0.
1
Isolate x2
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Inverse operations will be used to isolate x2. Here, 500 will be added to both sides of the equation. Then, the left- and right-hand sides will be divided by 5.
5x2500=0
5x2=500
x2=100
2
Take Square Roots
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Now that x2 has been isolated, it is necessary to undo the exponent. Exponents and radicals of the same index undo each other. Therefore, square roots undo powers of 2.
x2=100

x=±10
Note that the negative solution is also considered along with the principal root when solving the equation.

Method

Factoring a Quadratic Trinomial

When trying to factor a quadratic trinomial of the form ax2+bx+c, it can be difficult to see its factors. Consider the following expression.
Here, a=8, b=26, and c=6. There are six steps to factor this trinomial.
1
Factor Out the GCF of a, b, and c
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To fully factor a quadratic trinomial, the Greatest Common Factor (GCF) of a, b, and c has to be factored out first. To identify the GCF of these numbers, their prime factors will be listed.
It can be seen that 8, 26, and 6 share exactly one factor, 2.
Now, 2 can be factored out.
8x2+26x+6
2(4)x2+2(13)x+2(3)
2(4x2+13x+3)
In the remaining steps, the factored coefficient 2 before the parentheses can be ignored. The new considered quadratic trinomial is 4x2+13x+3. Therefore, the current values of a, b, and c, are 4, 13, and 3, respectively. If the GCF of the coefficients is 1, this step can be ignored.
2
Find the Factor Pair of ac Whose Sum Is b
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It is known that a=4 and c=3, so ac=12>0. Therefore, the factors must have the same sign. Also, b=13. Since the sum of the factors is positive and they must have the same sign, both factors must be positive. All positive factor pairs of 12 can now be listed and their sums checked.

Factors of ac Sum of Factors
1 and 12
2 and 6
3 and 4

In this case, the correct factor pair is 1 and 12. The following table sums up how to determine the signs of the factors based on the values of ac and b.

ac b Factors
Positive Positive Both positive
Positive Negative Both negative
Negative Positive One positive and one negative. The absolute value of the positive factor is greater.
Negative Negative One positive and one negative. The absolute value of the negative factor is greater.

Such analysis makes the list of possible factor pairs shorter.

3
Write bx as a Sum
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The factor pair obtained in the previous step will be used to rewrite the x-term — the linear term — of the quadratic trinomial as a sum. Remember that the factors are 1 and 12.
13x
12x+1x
12x+x
The linear term 13x can be rewritten in the original expression as 12x+x.
4
Factor Out the GCF of the First Two Terms
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The expression has four terms, which can be grouped into the and the last two terms. Then, the GCF of each group can be factored out.
The first two terms, 4x2 and 12x, can be factored.
The GCF of 4x2 and 12x is
4xx+12x+x+3
4xx+4x3+x+3
4x(x+3)+x+3
5
Factor Out the GCF of the Last Two Terms
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The process used in Step 3 will be repeated for the last two terms. In this case, x and 3 cannot be factored, so their GCF is 1.
4x(x+3)+x+3
4x(x+3)+1x+3
4x(x+3)+1x+13
4x(x+3)+1(x+3)
6
Factor Out the Common Factor
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If all the previous steps have been performed correctly, there should now be two terms with a common factor.
The common factor will be factored out.
4x(x+3)+1(x+3)
(4x+1)(x+3)
The factored form of 4x2+13x+3 is (4x+1)(x+3). Remember that the original trinomial was 8x2+26x+6 and that the GCF 2 was factored out in Step 1. This GCF has to be included in the final result.

Example

Solve the quadratic equation by factoring

fullscreen
Solve the following equation by factoring.
x22x3=0
Show Solution expand_more
We'll focus on the left-hand side of the equation before solving for x. Notice that 1 and -3 are the factors that multiply to equal -3 and add to -2.
x22x3=0
x2+x3x3=0
x(x+1)3x3=0
x(x+1)3(x+1)=0
(x+1)(x3)=0
Notice that, in factored form, the numbers in the parentheses with x are the numbers from the chosen factor pair. This is because the coefficient of x2 is 1. We could have written the expression in factored form immediately after finding the factor pair. Lastly, we can use the zero product property to solve the equation.
Thus, the solutions to the equation x22x3=0 are x=-1 and x=3.
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