Other Types of Equations

So far, we have considered only linear and quadratic equations. In this section, we study the equations that are not either linear or quadratic but can be transferred to one of them and then can be solved.

Equations Reducible to Quadratic Form

Many equations, while not quadratic themselves, can be transformed into a quadratic form. An equation is in quadratic form if it can be written as

where represents some expression involving . If the solutions to are and , then this equation is equivalent to the two equations

Therefore, to solve the original equation in terms of , we would then solve these two equations for . This idea is central to solving all the examples in this section.

Biquadratic Equation Solve the equation .

Solution

Notice that this is a biquadratic equation because it only has even powers of

. If we substitute

, then the equation becomes

We can solve this quadratic equation for

using factoring or the quadratic formula. Using the quadratic formula, we get

which means that

. Since

, we have

Since the square of a real number cannot be negative, the equation

has no real solutions. Taking the square root of both sides of the first equation gives us

Therefore, the real solutions of

are

and

Equation with Fractional Exponents

Solve the equation .

Solution

To eliminate the negative exponent, we multiply both sides of the equation by

, obtaining

This equation is quadratic in

, so if we let

we obtain:

This can be solved by factoring. To do this, we are looking for two numbers that multiply to

and add up to

. These two numbers are

and

, so we have:

Thus, either

. Solving for u we obtain

Since

, we have

Taking the cube root of each side gives us

Equation with a Repeated Quadratic Expression

Solve the equation .

Solution

Notice that the expression

appears in both factors on the left-hand side. If we let

, we obtain

Expanding the product, we have

which gives us

To factor the quadratic expression, we need to find two numbers that multiply to 14 and add up to 9. Those numbers are 7 and 2, so we have

Therefore,

which means

Since

, we have

Solving the first quadratic equation:

To factor the expression we look for two numbers that multiply to 2 and add up to 3. Those numbers are 1 and 2, so we have

which has roots

and

. Solving the second quadratic equation:

using the quadratic formula:

This means that this equation has no real solutions. Therefore, the solutions of the given equations are

and

Equation with Four Linear Factors

Solve the equation .

Solution

When multiplying the first and fourth factors together, we obtain

while when we multiply the second and third factors we obtain

Letting

the original equation can be written as

Expanding the product gives us

To factor this, we need to find two numbers that multiply to -96 and add up to 10. Those numbers are 16 and -6 so we obtain:

Therefore,

, which means that

. Since

, we have

The second equation gives us

To factor this, we are looking for two numbers that multiply to -6 and add up to 5, which are 6 and -1. So we have:

which has roots

. The first equation gives us

Using the quadratic formula, we have

Thus the roots are

, and

Equation using Completing the Square

Solve the equation .

Solution

To make the equation more convenient to solve, we will use the method of completing the square. By completing the square for the first two terms, we obtain

Now we set

so that the above equation becomes

This can be factored to obtain:

This means that

, so we have

Since

, we obtain

Solving the first equation, we have

Using the quadratic formula

Solving the second equation, we have

Using the quadratic formula:

Thus the solutions of the given equation are

and

Equation Involving a Reciprocal Expression

Solve the equation

Solution

Notice that the second fraction is the reciprocal of the first. Let

. Then the equation becomes

Multiplying both sides by

, we obtain

Factoring this quadratic equation:

We look for two numbers whose sum is

and whose product is

. These numbers must both be negative, as their sum is negative and their product is positive. The numbers are

and

. Therefore,

Thus, either

, which means

, or

, which means

. Therefore, we have two equations:

From the first equation, we get

which simplifies to

The second equation gives

which simplifies to

yielding

Factoring:

We look for two numbers whose sum is

and whose product is

. The numbers are

and

. Thus,

Thus,

Considering both (*) and (**), the solutions are

All the values of

thus found are solutions of the given equation since they cause none of its denominators to vanish.

An Equation Involving Fractional Powers

Find all solutions of the equation .

Solution

This equation can be solved using a substitution to turn it into a quadratic equation. Since

, we can let

. This means that

. Using this substitution, the equation becomes

We can factor the quadratic expression in

as follows. We need to find two numbers that multiply to -2 and add to 1. Those two numbers are 2 and -1. This means

Therefore, either

, which gives us

Since

, we have

To find

, we raise both sides of each equation to the power of 6. For the first equation,

which gives us

For the second equation

which gives us

However, we need to check if these solutions satisfy the original equation, as these steps might introduce spurious solutions. If we plug in

in the original equation we obtain:

is not a solution of the original equation. If we plug in

we obtain:

is a solution. Therefore, the only solution is

Equations Involving Fractional Expressions

To solve equations involving fractional expressions, eliminate the denominators, for example, by multiplying each side by the ==least common multiple== (LCM) of all denominators --- although any common multiple works. Then test all the solutions to find the extraneous ones.

Solve the equation .

Solution

To eliminate the denominators, we multiply each side by the least common multiple of all denominators:

. Because

, the LCM of the denominators is

Using the quadratic formula, we find the solutions of the above equation:

. Now we need to check if these values satisfy the original equations:

Checking :

Because LHS = RHS, is a solution.

Checking :

so is not acceptable and the only solution is .

Note that in the above example, the expression that we multiplied both sides to does not vanish when . So could not be an extraneous solution and we did not have to test whether or not it satisfies the original equation.

Solve

Solution

Clearing of fractions (by multiplying each term by

) and simplifying, we get:

Using the quadratic formula, we have:

Thus, the solutions are

and

. Both of these values of

are valid roots of the given equation since none of the denominators (

) vanish when

takes these values.

Solve

Solution

We can factor the denominators to get

The lowest common denominator is

. Multiplying through by this, we obtain:

Factoring this quadratic, we get

. However,

makes the denominators of the first two terms zero. Therefore,

is not a solution. Hence, the only solution is

Radical Equations

Radical equations are equations in which the variable (the unknown) is under the square root, cubic root, or higher root, like

To solve radical equations:

Isolate the most complicated radical term on one side and move the rest of terms to the other side.
If the radical is a square root, square both side. If the radical is a cubic root, cube both sides and in general, for an th root radical, raise both sides to the th power. Then simplify the equation.
Repeat Steps 1 and 2 with the effort to eliminate all radicals involving the unknown.
Solve the resulting equation.
Test each solution by substitution in the original equation and determine which solutions satisfy the original equation.

Raising both sides of an equation to an even power may introduce extraneous solutions.
Recall that if is positive, represents only the positive root of . Similarly where is even represents only the positive th root.

Solve each equation:

(a)

(b)

Solution

(a) Rewrite the equation as

Then we square both sides:

Therefore

. Substituting

in the original equation, we get

Because LHS

RHS,

is an extraneous solution.

Substituting in the original equation, we get

and LHS RHS. So the only solution of this equation is .

(b) Rewrite the equation as

or

which gives or . Substituting in the original equation, we get

Because LHS RHS, is a solution. Substituting in the original equation, we get

Because LHS RHS, is not a solution (it is an extraneous solution) and so is the only solution.

Solve each equation:

(a)

(b)

Solution

(a)

Let's check if satisfies the original equation:

Because LHS RHS, is the solution.

(b)

Now we can factor it by trial and error as

which gives or . Alternatively we can use the quadratic equation to find the roots

which gives or . We have to test these values to see if they satisfy the original equation.

Substituting 2 for :

Because LHS RHS, is not a solution.

Substituting 38 for :

Because LHS RHS, is not a solution. Therefore, this problem does not have any solutions.

Other Types of Equations

Other Types of Equations

Equations Reducible to Quadratic Form

Equations Involving Fractional Expressions

Radical Equations

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