# Chapter 1 Equations, Inequalities, and Mathematical Models

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Chapter 1 Equations, Inequalities, and Mathematical Models
1.6 Other Types of Equations

Objectives At the end of this session , you will be able to:
Solve polynomial equations by factoring. Solve radical equations. Solve equations with rational exponents. Solve equations that are quadratic in form. Solve equations involving absolute values.

Contents Other Types of Equations Polynomial Equations

1. Other Types of Equations
In this section, we will learn to solve certain types of equations such as polynomial equations, equations involving rational exponents, equations involving absolute values, and radical equations. Polynomial Equations: Let us first recall the definition of polynomials: Recall: Polynomial: A polynomial is a single term, or the sum of two or more terms, containing variables with whole number exponents. We now define Polynomial Equations as follows: A polynomial equation is one polynomial set equal to another polynomial. The linear and quadratic equations that we studied in the previous sections of this chapter can be thought of as polynomial equations of degrees 1 and 2. For example, 3x2 + 5 = 5x is a polynomial equation of degree 2.    In addition, 7x2 – 10x = 8x3 – 5 is a polynomial equation of degree 3 because 3 is the largest exponent. Similarly, 5x4 = 3x3 + 4x2 is a polynomial equation of degree 4.

This is the given equation in standard form.
2. Polynomial Equations Standard Form of a Polynomial Equation: The standard form of a polynomial equation is an xn + an-1 xn-1 + … + a x + a0 = 0. In a polynomial equation, move all terms to one side, thereby obtaining a zero on the other side. Arrange the terms in a descending order. Such an equation is said to be in the standard form. For example, 9x3 – 5x2 + 4x = 0 is a polynomial equation in standard form. The general form of the quadratic equation ax2 + bx + c = 0 is also an example of polynomial equation in standard form. Solving a Polynomial Equations by Factoring: These steps are followed while solving a polynomial equation by factoring: Step 1: Write the equation in the standard form. If the given equation is not in the standard from, move all the terms to one side and obtain a zero on the other side. Make sure the equation is written in the descending order of terms. For instance, we cannot solve an equation of the form x3 – 3x2 = 16x - 48. We need to change it to the standard form to solve the given equation. x3 – 3x2 - 16x + 48 = 16x - 48 – 16x (Subtract 16x and add 48 to both sides) x3 – 3x2 - 16x + 48 = 0 (Simplify) This is the given equation in standard form.

2. Polynomial Equations (Cont…)
Step 2: Factor Factor the given equation. x3 – 3x2 - 16x + 48 = 0  x2 (x - 3) – 16 (x - 3) = 0 (Taking out the common factors)  (x - 3) (x2 – 16) = 0 (Factoring by grouping)  (x - 3) (x2 – (4)2) = 0 (16 = (4)2)  (x - 3) (x - 4) (x + 4) = 0 (Using (a2 – b2) = (a - b) (a + b)) Step 3: Use the zero product principle. Set each of the factor equal to zero, using the zero product principle. Recall: The Zero Product Principle: If the product of two algebraic expressions is zero, then at least one of the factors is equal to zero. That is, if A . B = 0, then A = 0 or B = 0.  (x - 3) = or (x - 4) = 0 or (x + 4) = 0 Common Factor is x2. Common Factor is -16.

2. Polynomial Equations (Cont…)
Step 4: Solve the equations obtained in step 3. (x - 3) = or (x - 4) = 0 or (x + 4) = 0  x = or x – = or x + 4 – 4 = 0 - 4  x = or x = or x = -4 Thus, the proposed solution set for the given polynomial equation is {-4, 3, 4}. Step 5: Check the solutions in the original equation. Check the three solutions, -4, 3, and 4 by substituting them in the original equation. Check: Check: 3 Check: 4 x3 – 3x2 – 16x + 48 = x3 – 3x2 - 16x + 48 = 0 x3 – 3x2 - 16x + 48 = 0 (– 4)3 – 3(– 4)2 – 16(–4) (3)3 – 3(3)2 – 16(3) (4)3 – 3(4)2 – 16(4) – 64 – – 27 – – 48 – 0 = 0 (True) 0 = 0 (True) 0 = 0 (True) Therefore, the solution set for the given polynomial equation is {-4, 3, 4}. NOTE: The steps used for solving a polynomial equation are essentially the same as used for solving a quadratic equation by factoring. ? =

Step 2: Remove the radical sign. To remove the radical sign, raise each side of the equation to a power equal to the index of the root. For the previous example, we square both sides of equation (1) to remove the radical sign. Step 3: Solve the equation obtained in step 2.

Thus, the proposed solution set is {2, 6}. Step 4: Check for extraneous solutions. When solving a radical equation, extra solutions may be introduced when we raise both sides of the equation to an even power. Such solutions, which are not the solutions of the given equation, are called extraneous solutions. In other words, if a value is an extraneous solution, it is not a solution to the original problem. Let us check the proposed solutions by substituting both proposed solutions in the original equation. Check 2: Check 6: Therefore, 6 is an extraneous solution and the only solution to this radical equation is 2.

Now let us solve another example involving more than one radical expression. Example: Step 1: Isolate one of the radicals on one side of the equation. Step 2: Remove the radical sign. We remove the radical sign by squaring both sides of the equation.

Step 3: Repeat step 1 and 2 if there is a radical left in the equation obtained in step 2. As there is a radical expression left in the equation obtained in step 2, we repeat steps 1 and 2. Step 4: Solve the equation obtained in step 3. This is an additional step

Step 4: Check for extraneous solutions. Check the proposed solution in the original equation. Check -1: Check 3: Therefore, the solution set for the given equation is {-1. 3}. NOTE: While solving a radical equation, the following points need to kept in mind: Square the sides of the equation, not the terms. Consider, = = 5 (2 + 3)2 = (5)2 but (2)2 + (3)2 = (5)2 25 =  25 Always check the answer for the extraneous solutions.

Solving Radical Equations of the form xm/n = k: (Assume that m and n are positive integers, m/n is in lowest terms and k is a real number.) We have already learned that a radical expression can be expressed as (b)1/n using rational exponents. Let us learn how to solve equations involving rational exponents. Step 1: Isolate the base with the rational exponent. Consider the equation 2x5/3 – 10 = 0 Step 2: Remove the rational exponent. To remove the rational exponent, raise the expression that has a rational exponent to the reciprocal of that rational exponent That is, raise both sides of the equation xm/n = k to the n/m power. If m is even: If m is odd: xm/n = k xm/n = k (xm/n)n/m =  kn/m (xm/n)n/m = kn/m x =  kn/m x = kn/m NOTE: An odd index has only one root. Do not insert the  symbol when the numerator of the exponent is odd. Make sure you apply the same operations on either side of the equation to keep it balanced.

For the given example the numerator of the exponent is odd. Thus, we have: Step 3: Solve the equation obtained in step 2. In this example the equation that resulted from raising both sides to the 3/5 power is a linear equation. Also observe that it is already solved for x. Thus, the proposed solution is Step 4: Check for extraneous solutions. Check to find if is an extraneous solution by substituting the value in the original equation.

Some equations are not quadratic equations, but can be written in the quadratic form by using appropriate substitution. An equation is said to be in a quadratic form if it can be written in the form at2 + bt + c = 0, where t is an algebraic expression and a  0. For example, x4 – 3x2 – 8 = 0 is not a quadratic equation, but it can be written in the quadratic form by substituting x2 = t. x4 – 3x2 – 8 =  (x2)2 – 3x2 – 8 = 0 Substituting x2 = t, we get t2 – 3t – 8 = 0. This is a quadratic equation in t. Now we can solve this equation for t. Here are some more examples: Given Equation Substitution New Equation 3x4 + 5x2 + 7 = 0 x2 = t 3t2 + 5t + 7 = 0 2x2/3 + 4x1/3 + 6 = 0 Or 2(x1/3)2 + 4x1/3 + 6 = 0 x1/3 = t 2t2 + 4t + 6 = 0 (x2 - x)2 + 5(x2 - x) + 9 =0 (x2 - x) = t t2 + 5t + 9 = 0 t2 + 5t + 6 = 0

4. Equations That are Quadratic in Form (Cont…)
Solving an equation which is quadratic in form: Follow these steps for solving an equation which is quadratic in form: Step 1: Convert the given equation to quadratic form by making the appropriate substitution. Consider the equation Assume t = x1/3 . Rewriting the above equation as a quadratic equation in t, we get: 3t2 – 11t – 4 = 0 Step 2: Solve the quadratic equation obtained in step 1. 3t2 – 11t – 4 = 0 (Quadratic equation in t) 3t2 – 12t + t – 4 = 0 (Factor the quadratic equation) 3t(t - 4) + (t - 4) = 0 (t - 4) (3t + 1) = 0 (t - 4) = 0 or (3t + 1) = 0 (Apply the zero product principle) t – = t + 1 – 1 = 0 – 1 t = t = -1 t = -1/3

4. Equations in Quadratic Form (Cont…)
Step 3: Replace the substitution by the original expression. t = 4 t = -1/3 (Solutions obtained in step 2) x1/3 = 4 x1/3 =-1/3 (Replace t with x1/3) Step 4: Solve for x. x1/3 = 4 x1/3 =-1/3 (x1/3)3 = (4)3/1 (x1/3)3 = (-1/3)3/1 x = 64 x = -1/27 Step 5: Check the proposed solution for an extraneous solution. Substitute the proposed solutions in the original equation.

5. Absolute Value Equation
Recall: Absolute Value: The absolute value measures the distance a number is away from the origin (zero) on the number line. The absolute value is always positive. The absolute value of a number x is given by: |x| = Solving an Absolute Value Equation: The steps to be followed for solving an absolute value equations are as follows: Step 1: Use the definition of absolute value to set up the equation without absolute values. Consider the equation |7x + 2| = 10, using the definition of absolute value, we get: 7x + 2 = 10 or 7x + 2 = -10   If |x| = d, then No solution where d < 0  This is because the distance (d)  cannot be negative. If |x| = d, then x = d    or    x = -d where d > 0 (Two equations are set up)

Absolute Value Equation (Cont…)
Step 2: Solve the equations set up in step 1. 7x + 2 = 10 or 7x + 2 = -10 7x + 2 – 2 = 10 – 2 or 7x + 2 – 2 = -10 – 2 7x = 8 or 7x = -12 7x/7 = 8/7 or 7x/7 = -12/7 x = 8/7 or x = -12/7 Step 3: Check the proposed solution by substituting the values in the original equation. Check 8/7: Check –12/7: |7x + 2| = |7x + 2| = 10 |7. 8/7 + 2| = 10   |7. –12/7 + 2| = 10 |8 + 2| = | | = 10 |10| = |-10| = 10 10 = 10 (True) = 10 (True) Thus, the solution set for the given equation is {-12/7, 8/7}.

Absolute Value Equation (Cont…)
Let us consider another example. Example: Solve 3|2x – 3| - 8 = 25 To solve such absolute value equations, first isolate the expression containing the absolute value symbol. 3|2x – 3| = (Add 8 to both sides) 3|2x – 3| = 33 (Simplify) |2x – 3| = 33/3 (Divide both sides by 3) |2x – 3| = 11 Now this equation can be solved as explained in the previous example. The solution set for this equation is {-4, 7}. Example: Solve |10x - 5| = -5 In this example, the absolute value is set equal to a negative number. There is no value of x that we can substitute that will be a solution because when we take the absolute value of the left side it will always be positive or zero, never negative. Thus, the answer for this example is no solution. The solution set is , the empty set.

6. Summary Let us recall what we have learned so far:
Polynomial equations with degree 3 or greater can be solved by factoring and using the zero product principle. A radical equation is an equation in which the variable occurs under the radical sign. A radical equation can be solved by isolating the radical and raising both sides of the equation to a power equal to the radical’s index. A radical equation with rational exponents can be solved by isolating the expression with the rational exponent and raising both sides of the equation to a power that is a reciprocal of the rational exponent. An equation is said to be in a quadratic form if it can be written in the form at2 + bt + c = 0, where t is an algebraic expression and a  0. Absolute value equations are equations of the form |X| = c, c > 0.