# 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

${t}^{25}-{t}^{5}+1=0$

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

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

## 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

what is the coefficient of -4×
-1
Shedrak
the operation * is x * y =x + y/ 1+(x × y) show if the operation is commutative if x × y is not equal to -1
An investment account was opened with an initial deposit of \$9,600 and earns 7.4% interest, compounded continuously. How much will the account be worth after 15 years?
lim x to infinity e^1-e^-1/log(1+x)
given eccentricity and a point find the equiation
12, 17, 22.... 25th term
12, 17, 22.... 25th term
Akash
College algebra is really hard?
Absolutely, for me. My problems with math started in First grade...involving a nun Sister Anastasia, bad vision, talking & getting expelled from Catholic school. When it comes to math I just can't focus and all I can hear is our family silverware banging and clanging on the pink Formica table.
Carole
I'm 13 and I understand it great
AJ
I am 1 year old but I can do it! 1+1=2 proof very hard for me though.
Atone
hi
Not really they are just easy concepts which can be understood if you have great basics. I am 14 I understood them easily.
Vedant
find the 15th term of the geometric sequince whose first is 18 and last term of 387
I know this work
salma
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
hmm well what is the answer
Abhi
If f(x) = x-2 then, f(3) when 5f(x+1) 5((3-2)+1) 5(1+1) 5(2) 10
Augustine
how do they get the third part x = (32)5/4
make 5/4 into a mixed number, make that a decimal, and then multiply 32 by the decimal 5/4 turns out to be
AJ
how
Sheref
can someone help me with some logarithmic and exponential equations.
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
hmm
Abhi
is it a question of log
Abhi
🤔.
Abhi
I rally confuse this number And equations too I need exactly help
salma
But this is not salma it's Faiza live in lousvile Ky I garbage this so I am going collage with JCTC that the of the collage thank you my friends
salma
Commplementary angles
hello
Sherica
im all ears I need to learn
Sherica
right! what he said ⤴⤴⤴
Tamia
hii
Uday
hi
salma
hi
Ayuba
Hello
opoku
hi
Ali
greetings from Iran
Ali
salut. from Algeria
Bach
hi
Nharnhar
what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
a perfect square v²+2v+_
kkk nice