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2.6 Other types of equations  (Page 2/10)

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Solve:   ( x + 5 ) 3 2 = 8.

{ −1 }

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Solving equations using factoring

We have used factoring to solve quadratic equations, but it is a technique that we can use with many types of polynomial equations, which are equations that contain a string of terms including numerical coefficients and variables. When we are faced with an equation containing polynomials of degree higher than 2, we can often solve them by factoring.

Polynomial equations

A polynomial of degree n is an expression of the type

a n x n + a n − 1 x n − 1 +       ⋅     ⋅     ⋅     + a 2 x 2 + a 1 x + a 0

where n is a positive integer and   a n , … , a 0   are real numbers and   a n ≠ 0.

Setting the polynomial equal to zero gives a polynomial equation    . The total number of solutions (real and complex) to a polynomial equation is equal to the highest exponent n .

Solving a polynomial by factoring

Solve the polynomial by factoring:   5 x 4 = 80 x 2 .

First, set the equation equal to zero. Then factor out what is common to both terms, the GCF.

5 x 4 − 80 x 2 = 0 5 x 2 ( x 2 − 16 ) = 0

Notice that we have the difference of squares in the factor   x 2 − 16 , which we will continue to factor and obtain two solutions. The first term,   5 x 2 , generates, technically, two solutions as the exponent is 2, but they are the same solution.

5 x 2 = 0 x = 0 x 2 − 16 = 0 ( x − 4 ) ( x + 4 ) = 0 x = 4 x = −4

The solutions are   0  (double solution), 4 , and   −4.

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Solve by factoring:   12 x 4 = 3 x 2 .

x = 0 , x = 1 2 , x = − 1 2

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Solve a polynomial by grouping

Solve a polynomial by grouping:   x 3 + x 2 − 9 x − 9 = 0.

This polynomial consists of 4 terms, which we can solve by grouping. Grouping procedures require factoring the first two terms and then factoring the last two terms. If the factors in the parentheses are identical, we can continue the process and solve, unless more factoring is suggested.

x 3 + x 2 − 9 x − 9 = 0 x 2 ( x + 1 ) − 9 ( x + 1 ) = 0 ( x 2 − 9 ) ( x + 1 ) = 0

The grouping process ends here, as we can factor   x 2 − 9   using the difference of squares formula.

( x 2 − 9 ) ( x + 1 ) = 0 ( x − 3 ) ( x + 3 ) ( x + 1 ) = 0 x = 3 x = −3 x = −1

The solutions are 3 , −3 , and   −1.   Note that the highest exponent is 3 and we obtained 3 solutions. We can see the solutions, the x- intercepts, on the graph in [link] .

Coordinate plane with the x-axis ranging from negative 5 to 5 and the y-axis ranging from negative 30 to 20 in intervals of 5. The function x cubed plus x squared minus nine times x minus nine equals zero is graphed along with the points (negative 3,0), (negative 1,0), and (3,0).
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Solving radical equations

Radical equations are equations that contain variables in the radicand    (the expression under a radical symbol), such as

3 x + 18 = x x + 3 = x − 3 x + 5 − x − 3 = 2

Radical equations may have one or more radical terms, and are solved by eliminating each radical, one at a time. We have to be careful when solving radical equations, as it is not unusual to find extraneous solutions    , roots that are not, in fact, solutions to the equation. These solutions are not due to a mistake in the solving method, but result from the process of raising both sides of an equation to a power. However, checking each answer in the original equation will confirm the true solutions.

Radical equations

An equation containing terms with a variable in the radicand is called a radical equation    .

Given a radical equation, solve it.

  1. Isolate the radical expression on one side of the equal sign. Put all remaining terms on the other side.
  2. If the radical is a square root, then square both sides of the equation. If it is a cube root, then raise both sides of the equation to the third power. In other words, for an n th root radical, raise both sides to the n th power. Doing so eliminates the radical symbol.
  3. Solve the remaining equation.
  4. If a radical term still remains, repeat steps 1–2.
  5. Confirm solutions by substituting them into the original equation.

Questions & Answers

A golfer on a fairway is 70 m away from the green, which sits below the level of the fairway by 20 m. If the golfer hits the ball at an angle of 40° with an initial speed of 20 m/s, how close to the green does she come?
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2. A sled plus passenger with total mass 50 kg is pulled 20 m across the snow (0.20) at constant velocity by a force directed 25° above the horizontal. Calculate (a) the work of the applied force, (b) the work of friction, and (c) the total work.
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A string is 3.00 m long with a mass of 5.00 g. The string is held taut with a tension of 500.00 N applied to the string. A pulse is sent down the string. How long does it take the pulse to travel the 3.00 m of the string?
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Source:  OpenStax, Algebra and trigonometry. OpenStax CNX. Nov 14, 2016 Download for free at https://legacy.cnx.org/content/col11758/1.6
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