# 2.6 Other types of equations  (Page 5/10)

 Page 5 / 10

Solve the following rational equation: $\text{\hspace{0.17em}}\frac{-4x}{x-1}+\frac{4}{x+1}=\frac{-8}{{x}^{2}-1}.$

We want all denominators in factored form to find the LCD. Two of the denominators cannot be factored further. However, $\text{\hspace{0.17em}}{x}^{2}-1=\left(x+1\right)\left(x-1\right).\text{\hspace{0.17em}}$ Then, the LCD is $\text{\hspace{0.17em}}\left(x+1\right)\left(x-1\right).\text{\hspace{0.17em}}$ Next, we multiply the whole equation by the LCD.

$\begin{array}{ccc}\hfill \left(x+1\right)\left(x-1\right)\left[\frac{-4x}{x-1}+\frac{4}{x+1}\right]& =& \left[\frac{-8}{\left(x+1\right)\left(x-1\right)}\right]\left(x+1\right)\left(x-1\right)\hfill \\ \hfill -4x\left(x+1\right)+4\left(x-1\right)& =& -8\hfill \\ \hfill -4{x}^{2}-4x+4x-4& =& -8\hfill \\ \hfill -4{x}^{2}+4& =& 0\hfill \\ \hfill -4\left({x}^{2}-1\right)& =& 0\hfill \\ \hfill -4\left(x+1\right)\left(x-1\right)& =& 0\hfill \\ \hfill x& =& -1\hfill \\ \hfill x& =& 1\hfill \end{array}$

In this case, either solution produces a zero in the denominator in the original equation. Thus, there is no solution.

Solve $\text{\hspace{0.17em}}\frac{3x+2}{x-2}+\frac{1}{x}=\frac{-2}{{x}^{2}-2x}.$

$x=-1,$ $x=0$ is not a solution.

Access these online resources for additional instruction and practice with different types of equations.

## Key concepts

• Rational exponents can be rewritten several ways depending on what is most convenient for the problem. To solve, both sides of the equation are raised to a power that will render the exponent on the variable equal to 1. See [link] , [link] , and [link] .
• Factoring extends to higher-order polynomials when it involves factoring out the GCF or factoring by grouping. See [link] and [link] .
• We can solve radical equations by isolating the radical and raising both sides of the equation to a power that matches the index. See [link] and [link] .
• To solve absolute value equations, we need to write two equations, one for the positive value and one for the negative value. See [link] .
• Equations in quadratic form are easy to spot, as the exponent on the first term is double the exponent on the second term and the third term is a constant. We may also see a binomial in place of the single variable. We use substitution to solve. See [link] and [link] .

## Verbal

In a radical equation, what does it mean if a number is an extraneous solution?

This is not a solution to the radical equation, it is a value obtained from squaring both sides and thus changing the signs of an equation which has caused it not to be a solution in the original equation.

Explain why possible solutions must be checked in radical equations.

Your friend tries to calculate the value $\text{\hspace{0.17em}}-{9}^{\frac{3}{2}}$ and keeps getting an ERROR message. What mistake is he or she probably making?

He or she is probably trying to enter negative 9, but taking the square root of $\text{\hspace{0.17em}}-9\text{\hspace{0.17em}}$ is not a real number. The negative sign is in front of this, so your friend should be taking the square root of 9, cubing it, and then putting the negative sign in front, resulting in $\text{\hspace{0.17em}}-27.$

Explain why $\text{\hspace{0.17em}}|2x+5|=-7\text{\hspace{0.17em}}$ has no solutions.

Explain how to change a rational exponent into the correct radical expression.

A rational exponent is a fraction: the denominator of the fraction is the root or index number and the numerator is the power to which it is raised.

## Algebraic

For the following exercises, solve the rational exponent equation. Use factoring where necessary.

${x}^{\frac{2}{3}}=16$

${x}^{\frac{3}{4}}=27$

$x=81$

$2{x}^{\frac{1}{2}}-{x}^{\frac{1}{4}}=0$

${\left(x-1\right)}^{\frac{3}{4}}=8$

$x=17$

${\left(x+1\right)}^{\frac{2}{3}}=4$

${x}^{\frac{2}{3}}-5{x}^{\frac{1}{3}}+6=0$

${x}^{\frac{7}{3}}-3{x}^{\frac{4}{3}}-4{x}^{\frac{1}{3}}=0$

For the following exercises, solve the following polynomial equations by grouping and factoring.

${x}^{3}+2{x}^{2}-x-2=0$

$x=-2,1,-1$

$3{x}^{3}-6{x}^{2}-27x+54=0$

$4{y}^{3}-9y=0$

${x}^{3}+3{x}^{2}-25x-75=0$

${m}^{3}+{m}^{2}-m-1=0$

$m=1,-1$

$2{x}^{5}-14{x}^{3}=0$

$5{x}^{3}+45x=2{x}^{2}+18$

$x=\frac{2}{5}$

For the following exercises, solve the radical equation. Be sure to check all solutions to eliminate extraneous solutions.

$\sqrt{3x-1}-2=0$

$\sqrt{x-7}=5$

$x=32$

$\sqrt{x-1}=x-7$

$\sqrt{3t+5}=7$

$t=\frac{44}{3}$

$\sqrt{t+1}+9=7$

$\sqrt{12-x}=x$

$x=3$

$\sqrt{2x+3}-\sqrt{x+2}=2$

$\sqrt{3x+7}+\sqrt{x+2}=1$

$x=-2$

$\sqrt{2x+3}-\sqrt{x+1}=1$

For the following exercises, solve the equation involving absolute value.

$|3x-4|=8$

$x=4,\frac{-4}{3}$

$|2x-3|=-2$

$|1-4x|-1=5$

$x=\frac{-5}{4},\frac{7}{4}$

$|4x+1|-3=6$

$|2x-1|-7=-2$

$x=3,-2$

$|2x+1|-2=-3$

$|x+5|=0$

$x=-5$

$-|2x+1|=-3$

For the following exercises, solve the equation by identifying the quadratic form. Use a substitute variable and find all real solutions by factoring.

${x}^{4}-10{x}^{2}+9=0$

$x=1,-1,3,-3$

$4{\left(t-1\right)}^{2}-9\left(t-1\right)=-2$

${\left({x}^{2}-1\right)}^{2}+\left({x}^{2}-1\right)-12=0$

$x=2,-2$

${\left(x+1\right)}^{2}-8\left(x+1\right)-9=0$

${\left(x-3\right)}^{2}-4=0$

$x=1,5$

## Extensions

For the following exercises, solve for the unknown variable.

${x}^{-2}-{x}^{-1}-12=0$

$\sqrt{{|x|}^{2}}=x$

All real numbers

${t}^{10}-{t}^{5}+1=0$

$|{x}^{2}+2x-36|=12$

$x=4,6,-6,-8$

## Real-world applications

For the following exercises, use the model for the period of a pendulum, $\text{\hspace{0.17em}}T,$ such that $\text{\hspace{0.17em}}T=2\pi \sqrt{\frac{L}{g}},$ where the length of the pendulum is L and the acceleration due to gravity is $\text{\hspace{0.17em}}g.$

If the acceleration due to gravity is 9.8 m/s 2 and the period equals 1 s, find the length to the nearest cm (100 cm = 1 m).

If the gravity is 32 ft/s 2 and the period equals 1 s, find the length to the nearest in. (12 in. = 1 ft). Round your answer to the nearest in.

10 in.

For the following exercises, use a model for body surface area, BSA, such that $\text{\hspace{0.17em}}BSA=\sqrt{\frac{wh}{3600}},$ where w = weight in kg and h = height in cm.

Find the height of a 72-kg female to the nearest cm whose $\text{\hspace{0.17em}}BSA=1.8.$

Find the weight of a 177-cm male to the nearest kg whose $\text{\hspace{0.17em}}BSA=2.1.$

90 kg

why {2kπ} union {kπ}={kπ}?
why is {2kπ} union {kπ}={kπ}? when k belong to integer
Huy
if 9 sin theta + 40 cos theta = 41,prove that:41 cos theta = 41
what is complex numbers
give me treganamentry question
Solve 2cos x + 3sin x = 0.5
madras university algebra questions papers first year B. SC. maths
Hey
Rightspect
hi
chesky
Give me algebra questions
Rightspect
how to send you
Vandna
What does this mean
cos(x+iy)=cos alpha+isinalpha prove that: sin⁴x=sin²alpha
cos(x+iy)=cos aplha+i sinalpha prove that: sinh⁴y=sin²alpha
rajan
cos(x+iy)=cos aplha+i sinalpha prove that: sinh⁴y=sin²alpha
rajan
is there any case that you can have a polynomials with a degree of four?
victor
***sscc.edu/home/jdavidso/math/catalog/polynomials/fourth/fourth.html
Oliver
can you solve it step b step
give me some important question in tregnamentry
Anshuman
what is linear equation with one unknown 2x+5=3
-4
Joel
x=-4
Joel
x=-1
Joan
I was wrong. I didn't move all constants to the right of the equation.
Joel
x=-1
Cristian
y=x+1
gary
what is the VA Ha D R X int Y int of f(x) =x²+4x+4/x+2 f(x) =x³-1/x-1
can I get help with this?
Wayne
Are they two separate problems or are the two functions a system?
Wilson
Also, is the first x squared in "x+4x+4"
Wilson
x^2+4x+4?
Wilson
thank you
Wilson
Wilson
f(x)=x square-root 2 +2x+1 how to solve this value
Wilson
what is algebra
The product of two is 32. Find a function that represents the sum of their squares.
Paul
if theta =30degree so COS2 theta = 1- 10 square theta upon 1 + tan squared theta
how to compute this 1. g(1-x) 2. f(x-2) 3. g (-x-/5) 4. f (x)- g (x)
hi
John
hi
Grace
what sup friend
John
not much For functions, there are two conditions for a function to be the inverse function:   1--- g(f(x)) = x for all x in the domain of f     2---f(g(x)) = x for all x in the domain of g Notice in both cases you will get back to the  element that you started with, namely, x.
Grace