Solving Equations Algebraically
Contents: This page corresponds to § 2.4 (p. 200) of the text.
Suggested Problems from Text:
p. 212 #7, 8, 11, 15, 17, 18, 23, 26, 35, 38, 41, 43, 46, 47, 51, 54, 57, 60, 63, 66, 71, 72, 75, 76, 81, 87, 88, 95, 97
Equations Involving Radicals
Polynomial Equations of Higher Degree
Equations Involving Fractional Expressions or Absolute Values
A quadratic equation is one of the form ax2 + bx + c = 0, where a, b, and c are numbers, and a is not equal to 0.
This approach to solving equations is based on the fact that if the product of two quantities is zero, then at least one of the quantities must be zero. In other words, if a*b = 0, then either a = 0, or b = 0, or both. For more on factoring polynomials, see the review section P.3 (p.26) of the text.
2x2 - 5x - 12 = 0.
(2x + 3)(x - 4) = 0.
2x + 3 = 0 or x - 4 = 0.
x = -3/2, or x = 4.
If x2 = k, then x = ± sqrt(k).
x2 - 9 = 0.
x2 = 9.
x = 3, or x = -3.
x2 + 7 = 0.
x2 = -7.
x = ± .
Note that = = , so the solutions are
x = ± , two complex numbers.
The idea behind completing the square is to rewrite the equation in a form that allows us to apply the square root principle.
x2 +6x - 1 = 0.
x2 +6x = 1.
x2 +6x + 9 = 1 + 9.
The 9 added to both sides came from squaring half the coefficient of x, (6/2)2 = 9. The reason for choosing this value is that now the left hand side of the equation is the square of a binomial (two term polynomial). That is why this procedure is called completing the square. [ The interested reader can see that this is true by considering (x + a)2 = x2 + 2ax + a2. To get "a" one need only divide the x-coefficient by 2. Thus, to complete the square for x2 + 2ax, one has to add a2.]
(x + 3)2 = 10.
Now we may apply the square root principle and then solve for x.
x = -3 ± sqrt(10).
2x2 + 6x - 5 = 0.
2x2 + 6x = 5.
The method of completing the square demonstrated in the previous example only works if the leading coefficient (coefficient of x2) is 1. In this example the leading coefficient is 2, but we can change that by dividing both sides of the equation by 2.
x2 + 3x = 5/2.
Now that the leading coefficient is 1, we take the coefficient of x, which is now 3, divide it by 2 and square, (3/2)2 = 9/4. This is the constant that we add to both sides to complete the square.
x2 + 3x + 9/4 = 5/2 + 9/4.
The left hand side is the square of (x + 3/2). [ Verify this!]
(x + 3/2)2 = 19/4.
Now we use the square root principle and solve for x.
x + 3/2 = ± sqrt(19/4) = ± sqrt(19)/2.
x = -3/2 ± sqrt(19)/2 = (-3 ± sqrt(19))/2
So far we have discussed three techniques for solving quadratic equations. Which is best? That depends on the problem and your personal preference. An equation that is in the right form to apply the square root principle may be rearranged and solved by factoring as we see in the next example.
x2 = 16.
x2 - 16 = 0.
(x + 4)(x - 4) = 0.
x = -4, or x = 4.
In some cases the equation can be solved by factoring, but the factorization is not obvious.
The method of completing the square will always work, even if the solutions are complex numbers, in which case we will take the square root of a negative number. Furthermore, the steps necessary to complete the square are always the same, so they can be applied to the general quadratic equation
ax2 + bx + c = 0.
The result of completing the square on this general equation is a formula for the solutions of the equation called the Quadratic Formula.
The solutions for the equation ax2 + bx + c = 0 are
We are saying that completing the square always works, and we have completed the square in the general case, where we have a,b, and c instead of numbers. So, to find the solutions for any quadratic equation, we write it in the standard form to find the values of a, b, and c, then substitute these values into the Quadratic Formula.
One consequence is that you never have to complete the square to find the solutions for a quadratic equation. However, the process of completing the square is important for other reasons, so you still need to know how to do it!
Examples using the Quadratic Formula:
2x2 + 6x - 5 = 0.
In this case, a = 2, b = 6, c = -5. Substituting these values in the Quadratic Formula yields
Notice that we solved this equation earlier by completing the square.
Note: There are two real solutions. In terms of graphs, there are two intercepts for the graph of the function f(x) = 2x2 + 6x - 5.
4x2 + 4x + 1 = 0
In this example a = 4, b = 4, and c = 1.
There are two things to notice about this example.
- There is only one solution. In terms of graphs, this means there is only one x-intercept.
- The solution simplified so that there is no square root involved. This means that the equation could have been solved by factoring. (All quadratic equations can be solved by factoring! What I mean is it could have been solved easily by factoring.)
4x2 + 4x + 1 = 0.
(2x + 1)2 = 0.
x = -1/2.
x2 + x + 1 = 0
a = 1, b = 1, c = 1
Note: There are no real solutions. In terms of graphs, there are no intercepts for the graph of the function f(x) = x2 + x + 1. Thus, the solutions are complex because the graph of y = x2 + x + 1 has no x-intercepts.
The expression under the radical in the Quadratic Formula, b2 - 4ac, is called the discriminant of the equation. The last three examples illustrate the three possibilities for quadratic equations.
1. Discriminant > 0. Two real solutions.
2. Discriminant = 0. One real solution.
3. Discriminant < 0. Two complex solutions.
Notes on checking solutions
None of the techniques introduced so far in this section can introduce extraneous solutions. (See example 3 from the Linear Equations and Modeling section.) However, it is still a good idea to check your solutions, because it is very easy to make careless errors while solving equations.
The algebraic method, which consists of substituting the number back into the equation and checking that the resulting statement is true, works well when the solution is "simple", but it is not very practical when the solution involves a radical.
For instance, in our next to last example, 4x2 + 4x + 1 = 0, we found one solution x = -1/2.
The algebraic check looks like
4(-1/2)2 +4(-1/2) + 1 = 0.
4(1/4) - 2 + 1 = 0.
1 - 2 + 1 = 0.
0 = 0. The solution checks.
In the example before that, 2x2 + 6x - 5 = 0, we found two real solutions, x = (-3 ± sqrt(19))/2. It is certainly possible to check this algebraically, but it is not very easy. In this case either a graphical check, or using a calculator for the algebraic check are faster.
First, find decimal approximations for the two proposed solutions.
(-3 + sqrt(19))/2 = 0.679449.
(-3 - sqrt(19))/2 = -3.679449.
Now use a graphing utility to graph y = 2x2 + 6x - 5, and trace the graph to find approximately where the x-intercepts are. If they are close to the values above, then you can be pretty sure you have the correct solutions. You can also insert the approximate solution into the equation to see if both sides of the equation give approximately the same values. However, you still need to be careful in your claim that your solution is correct, since it is not the exact solution.
Note that if you had started with the equation 2x2 + 6x - 5 = 0 and gone directly to the graphing utility to solve it, then you would not get the exact solutions, because they are irrational. However, having found (algebraically) two numbers that you think are solutions, if the graphing utility shows that intercepts are very close to the numbers you found, then you are probably right!
Solve the following quadratic equations.
(a) 3x2 -5x - 2 = 0. Answer
(b) (x + 1)2 = 3. Answer
(c) x2 = 3x + 2. Answer
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Equations with radicals can often be simplified by raising to the appropriate power, squaring if the radical is a square root, cubing for a cube root, etc. This operation can introduce extraneous roots, so all solutions must be checked.
If there is only one radical in the equation, then before raising to a power, you should arrange to have the radical term by itself on one side of the equation.
Now that we have isolated the radical term on the right side, we square both sides and solve the resulting equation for x.
x = 0
When we substitute x = 0 into the original equation we get the statement 0 = 2, which is not true!
So, x = 0 is not a solution.
x = 3
When we substitute x = 3 into the original equation, we get the statement 3 = 3. This is true, so x = 3 is a solution.
Solution: x = 3.
Note: The solution is the x-coordinate of the intersection point of the graphs of y = x and y = sqrt(x+1)+1.
Look at what would have happened if we had squared both sides of the equation before isolating the radical term.
This is worse than what we started with!
If there is more than one radical term in the equation, then in general, we cannot eliminate all radicals by raising to a power one time. However, we can decrease the number of radical terms by raising to a power.
If the equation involves more than one radical term, then we still want to isolate one radical on one side and raise to a power. Then we repeat that process.
Now square both sides of the equation.
This equation has only one radical term, so we have made progress! Now isolate the radical term and then square both sides again.
Substituting x = 5/4 into the original equation yields
sqrt(9/4) + sqrt(1/4) = 2.
3/2 + 1/2 = 2.
This statement is true, so x = 5/4 is a solution.
Note on checking solutions:
The algebraic check was easy to do in this case. However, the graphical check has the advantage of showing that there are no solutions that we have not found, at least within the scope of the viewing rectangle. The solution is the x-coordinate of the intersection point of the graphs of y = 2 and y = sqrt(x+1)+sqrt(x-1).
Solve the equation sqrt(x+2) + 2 = 2x. Answer
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We have seen that any degree two polynomial equation (quadratic equation) in one variable can be solved with the Quadratic Formula. Polynomial equations of degree greater than two are more complicated. When we encounter such a problem, then either the polynomial is of a special form which allows us to factor it, or we must approximate the solutions with a graphing utility.
One common special case is where there is no constant term. In this case we may factor out one or more powers of x to begin the problem.
2x3 + 3x2 -5x = 0.
x (2x2 + 3x -5) = 0.
Now we have a product of x and a quadratic polynomial equal to 0, so we have two simpler equations.
x = 0, or 2x2 + 3x -5 = 0.
The first equation is trivial to solve. x = 0 is the only solution. The second equation may be solved by factoring. Note: If we were unable to factor the quadratic in the second equation, then we could have resorted to using the Quadratic Formula. [Verify that you get the same results as below.]
x = 0, or (2x + 5)(x - 1) = 0.
So there are three solutions: x = 0, x = -5/2, x = 1.
Note: The solution is found from the intercepts of the graphs of f(x) = 2x3 + 3x2 -5x.
x3 -2x2 -9x +18 = 0.
The coefficient of x2 is -2 times that of x3, and the same relationship exists between the coefficients of the third and fourth terms. Group terms one and two, and also terms three and four.
x2 (x - 2) - 9 (x - 2) = 0.
These groups share the common factor (x - 2), so we can factor the left hand side of the equation.
(x - 2)(x2 - 9) = 0.
Whenever we find a product equal to zero, we obtain two simpler equations.
x - 2 = 0, or x2 - 9 = 0.
x = 2, or (x + 3)(x - 3) = 0.
So, there are three solutions, x = 2, x = -3, x = 3.
Note: These solutions are found from the intercepts of the graph of f(x) = x3 -2x2 -9x +18.
x4 - x2 - 12 = 0.
This polynomial is not quadratic, it has degree four. However, it can be thought of as quadratic in x2.
(x2) 2 -(x2) - 12 = 0.
It might help you to actually substitute z for x2.
z2 - z - 12 = 0 This is a quadratic equation in z.
(z - 4)(z + 3) = 0.
z = 4 or z = -3.
We are not done, because we need to find values of x that make the original equation true. Now replace z by x2 and solve the resulting equations.
x2 = 4.
x = 2, x = -2.
x2 = -3.
x = i , or x = -i.
So there are four solutions, two real and two complex.
Note: These solutions are found from the intercepts of the graph of f(x) = x4 - x2 - 12.
A graph of f(x) = x4 - x2 - 12 and a zoom showing its local extrema.
Solve the equation x4 - 5x2 + 4 = 0. Answer
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The least common denominator is x(x + 2), so we multiply both sides by this product.
This equation is quadratic. The Quadratic Formula yields the solutions
Checking is necessary because we multiplied both sides by a variable expression. Using a graphing utility we see that both of these solutions check. The solution is the x-coordinate of the intersection point of the graphs of y = 1 and y = 2/x-1/(x+2).
5 | x - 1 | = x + 11.
The key to solving an equation with absolute values is to remember that the quantity inside the absolute value bars could be positive or negative. We will have two separate equations representing the different possibilities, and all solutions must be checked.
Case 1. Suppose x - 1 >= 0. Then | x - 1 | = x - 1, so we have the equation
5(x - 1) = x + 11.
5x - 5 = x + 11.
4x = 16.
x = 4, and this solution checks because 5*3 = 4 + 11.
Case 2. Suppose x - 1 < 0. Then x - 1 is negative, so | x - 1 | = -(x - 1). This point often confuses students, because it looks as if we are saying that the absolute value of an expression is negative, but we are not. The expression (x - 1) is already negative, so -(x - 1) is positive.
Now our equation becomes
-5(x - 1) = x + 11.
-5x + 5 = x + 11.
-6x = 6.
x = -1, and this solution checks because 5*2 = -1 + 11.
If you use the Java Grapher to check graphically, note that abs() is absolute value, so you would graph
5*abs(x - 1) - x - 11 and look at x-intercepts, or you can find the solution as the x-coordinates of the intersection points of the graphs of y = x+11 and y = 5*abs(x-1).
(a) Solve the equation Answer
(b) Solve the equation | x - 2 | = 2 - x/3 Answer
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