Question

Which of the following statements are true about the given rational equation?
4/x+6 +1/x²=x+10/x³+6x²
Check all of the boxes that apply.
A. x = 1 is a solution.
B. x = 0 is an extraneous solution.=
C. x = –1 is a solution.
D. x = –6 is an extraneous solution.

Answer 1

Answer:

Correct choices are A and C

Step-by-step explanation:

Consider the equation

$\dfrac{4}{x+6}+\dfrac{1}{x^2}=\dfrac{x+10}{x^3+6x^2}.$

Note that $x\neq 0,\ x\neq-6.$

The left side of this equation is equal to

$\dfrac{4}{x+6}+\dfrac{1}{x^2}=\dfrac{4\cdot x^2+1\cdot (x+6)}{(x+6)x^2}=\dfrac{4x^2+x+6}{x^3+6x^2}.$

Since the left and the right sides of this equation have the same denominators, their numerators are equal

$4x^2+x+6=x+10.$

Then

$4x^2=10-6,\\ \\4x^2=4,\\ \\x^2=1,\\ \\x_1=1,\ x_2=-1.$

Both these numbers are the solutions of the given equation.

Answer 2

Answer:

A and C is correct.

Step-by-step explanation:

$\text{Rational Equation: }\frac{4}{x+6}+\frac{1}{x^2}=\frac{x+10}{x^3+6x^2}$

First we simplify this equation (-1 + x^2)/(x (6 + x))

$\dfrac{x^2-1}{x(x+6)}=0$

Solution set: x=1,-1

We have to check all option which is solution of above equation.

A)  x=1, we will put x=1 into equation both sides.

$\text{Left Side:} \Rightarrow \frac{4}{x+6}+\frac{1}{x^2}$

$\Rightarrow \frac{4}{1+6}+\frac{1}{1^2}$

$\Rightarrow \frac{11}{7}$

$\text{Right Side:} \Rightarrow \frac{x+10}{x^3+6x^2}$

$\Rightarrow \frac{1+10}{1^3+6\times 1^2}$

$\Rightarrow \frac{11}{7}$

$\text{Left Side}=\text{Right Side}=\frac{11}{7}$

Thus, x=1 is a solution.

C)  x=-1, we will put x=-1 into equation both sides.

$\text{Left Side:} \Rightarrow \frac{4}{x+6}+\frac{1}{x^2}$

$\Rightarrow \frac{4}{-1+6}+\frac{1}{(-1)^2}$

$\Rightarrow \frac{9}{5}$

$\text{Right Side:} \Rightarrow \frac{-1+10}{-1^3+6(-1)^2}$

$\Rightarrow \frac{9}{-1+6\times 1}$

$\Rightarrow \frac{9}{5}$

$\text{Left Side}=\text{Right Side}= \frac{9}{5}$

Thus, x=-1 is a solution.

Thus, A and C is correct.