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

A and C is correct.

Explanation:

`\text{Rational Equation: }(4)/(x+6)+(1)/(x^2)=(x+10)/(x^3+6x^2)`

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

`(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 (4)/(x+6)+(1)/(x^2)`

`\Rightarrow (4)/(1+6)+(1)/(1^2)`

`\Rightarrow (11)/(7)`

`\text{Right Side:} \Rightarrow (x+10)/(x^3+6x^2)`

`\Rightarrow (1+10)/(1^3+6* 1^2)`

`\Rightarrow (11)/(7)`

`\text{Left Side}=\text{Right Side}=(11)/(7)`

Thus, x=1 is a solution.

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

`\text{Left Side:} \Rightarrow (4)/(x+6)+(1)/(x^2)`

`\Rightarrow (4)/(-1+6)+(1)/((-1)^2)`

`\Rightarrow (9)/(5)`

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

`\Rightarrow (9)/(-1+6* 1)`

`\Rightarrow (9)/(5)`

`\text{Left Side}=\text{Right Side}= (9)/(5)`

Thus, x=-1 is a solution.

Thus, A and C is correct.

Answer 2

Correct choices are A and C

Explanation:

Consider the equation

`(4)/(x+6)+(1)/(x^2)=(x+10)/(x^3+6x^2).`

Note that
`x\\eq 0,\ x\\eq-6.`

The left side of this equation is equal to

`(4)/(x+6)+(1)/(x^2)=(4\cdot x^2+1\cdot (x+6))/((x+6)x^2)=(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.