Solve the Rational Expression: Find x in (x³+1)/(x+1)²=x

Q: Why can't x = -1 be a solution even if it satisfies the quadratic?

Because when $ x = -1 $, the denominator $ (x+1)^2 = 0 $, which makes the fraction undefined. Division by zero is not allowed in mathematics!

Q: How do I know when to use the sum of cubes formula?

Look for expressions like $ x^3 + 1 $ or $ x^3 + 8 $. These factor as $ (x+1)(x^2-x+1) $ and $ (x+2)(x^2-2x+4) $ respectively. The formula is $ a^3 + b^3 = (a+b)(a^2-ab+b^2) $.

Q: What if I get multiple solutions from the quadratic formula?

Check each solution in the original equation! Some might make denominators zero (extraneous solutions) and must be rejected, while others are valid.

Q: Is there a faster way to solve this without cross-multiplying?

You could factor $ x^3 + 1 = (x+1)(x^2-x+1) $ first, then cancel $ (x+1) $ terms. But be careful - you still need to remember that $ x \neq -1 $!

Q: How do I verify my answer x = 1/2?

Substitute into the original equation: $ \frac{(\frac{1}{2})^3 + 1}{(\frac{1}{2}+1)^2} = \frac{\frac{1}{8} + 1}{(\frac{3}{2})^2} = \frac{\frac{9}{8}}{\frac{9}{4}} = \frac{1}{2} $ ✓

Step-by-step video solution

Watch the teacher solve the problem with clear explanations

00:00 Find X

00:03 Multiply by the denominator to eliminate the fraction

00:10 Note the domain of definition

00:14 The denominator must be different from 0

00:27 Use shortened multiplication formulas to expand the brackets

00:36 Properly expand brackets, multiply by each factor

00:52 Simplify what we can

01:02 Arrange the equation so one side equals 0

01:11 Use the root formula to find possible solutions

01:27 Calculate the square root of 9

01:32 This is one solution, but it's not valid due to the domain definition

01:42 Therefore this is the solution to the problem

Step-by-step written solution

Follow each step carefully to understand the complete solution

1

Understand the problem

Solve the following equation:

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

2

Step-by-step solution

To solve the equation $\frac{x^3 + 1}{(x+1)^2} = x$ , we will follow these steps:

Step 1: Set up the equation for solving by cross-multiplying.
Step 2: Simplify and solve the resulting polynomial equation.
Step 3: Solve for the values of $x$ .

Let's work through the solution:

Step 1: Cross-multiply to eliminate the fraction:

(x^3 + 1) = x \cdot (x+1)^2

Expand the right-hand side:

x \cdot (x^2 + 2x + 1) = x^3 + 2x^2 + x

Step 2: Set the expanded equation equal:

x^3 + 1 = x^3 + 2x^2 + x

Cancel $x^3$ from both sides:

1 = 2x^2 + x

Re-arrange the equation to form a standard quadratic equation:

0 = 2x^2 + x - 1

Step 3: Solve the quadratic equation using the quadratic formula:

\text{Here, } a = 2, \, b = 1, \, c = -1.

The quadratic formula is:

x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Substitute the values of $a$ , $b$ , and $c$ into the formula:

x = \frac{-1 \pm \sqrt{1^2 - 4 \times 2 \times (-1)}}{2 \times 2}

Calculate the discriminant and simplify:

x = \frac{-1 \pm \sqrt{1 + 8}}{4} = \frac{-1 \pm \sqrt{9}}{4}

Simplify further:

x = \frac{-1 \pm 3}{4}

This gives the solutions:

x = \frac{2}{4} = \frac{1}{2}

x = \frac{-4}{4} = -1

Since $x = -1$ would make the denominator zero, it is not allowed as a solution. Thus, the only valid solution is:

Therefore, the solution to the equation is $x = \frac{1}{2}$ .

3

Final Answer

$x=\frac{1}{2}$

Key Points to Remember

Essential concepts to master this topic

Cross-multiply: Eliminate fractions by multiplying both sides by denominator
Factor: Use $x^3 + 1 = (x+1)(x^2-x+1)$ to simplify
Check domain: Verify $x \neq -1$ since denominator cannot equal zero ✓

Common Mistakes

Avoid these frequent errors

Forgetting to check domain restrictions
Don't solve the equation and accept x = -1 as valid = division by zero error! When x = -1, the denominator (x+1)² becomes zero, making the original equation undefined. Always check that your solutions don't make any denominator equal to zero.

Practice Quiz

Test your knowledge with interactive questions

Determine if the simplification shown below is correct:

$\frac{7}{7\cdot8}=8$

FAQ

Everything you need to know about this question

Why can't x = -1 be a solution even if it satisfies the quadratic?

+

Because when $x = -1$ , the denominator $(x+1)^2 = 0$ , which makes the fraction undefined. Division by zero is not allowed in mathematics!

How do I know when to use the sum of cubes formula?

+

Look for expressions like $x^3 + 1$ or $x^3 + 8$ . These factor as $(x+1)(x^2-x+1)$ and $(x+2)(x^2-2x+4)$ respectively. The formula is $a^3 + b^3 = (a+b)(a^2-ab+b^2)$ .

What if I get multiple solutions from the quadratic formula?

+

Check each solution in the original equation! Some might make denominators zero (extraneous solutions) and must be rejected, while others are valid.

Is there a faster way to solve this without cross-multiplying?

+

You could factor $x^3 + 1 = (x+1)(x^2-x+1)$ first, then cancel $(x+1)$ terms. But be careful - you still need to remember that $x \neq -1$ !

How do I verify my answer x = 1/2?

+

Substitute into the original equation: $\frac{(\frac{1}{2})^3 + 1}{(\frac{1}{2}+1)^2} = \frac{\frac{1}{8} + 1}{(\frac{3}{2})^2} = \frac{\frac{9}{8}}{\frac{9}{4}} = \frac{1}{2}$ ✓

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Solve the Rational Expression: Find x in (x³+1)/(x+1)²=x

Rational Equations with Domain Restrictions

❤️ Continue Your Math Journey!

Step-by-step video solution

Step-by-step written solution

Understand the problem

Step-by-step solution

Final Answer

Key Points to Remember

Common Mistakes

Practice Quiz

FAQ

Why can't x = -1 be a solution even if it satisfies the quadratic?

How do I know when to use the sum of cubes formula?

What if I get multiple solutions from the quadratic formula?

Is there a faster way to solve this without cross-multiplying?

How do I verify my answer x = 1/2?

🌟 Unlock Your Math Potential

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