Solve the Rational Equation: (x³+1)/(x-1)² = x+4

Rational Equations with Polynomial Multiplication

Solve the following equation:

x3+1(x1)2=x+4 \frac{x^3+1}{(x-1)^2}=x+4

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations
00:00 Solve
00:08 Multiply to eliminate the fraction
00:24 Use the distributive property of multiplication to expand the parentheses
00:47 Calculate the multiplication
00:55 Make sure to properly open parentheses
01:04 Each term in the first parentheses multiplies each term in the second parentheses
01:25 Reduce where possible
01:32 Collect like terms
01:42 Arrange the equation so that one side equals 0
01:50 Use the quadratic formula to find possible solutions
02:05 Calculate the multiplication
02:20 Solve for the positive option
02:26 This is one solution
02:29 Now solve for the negative option
02:33 This is the second solution
02:37 Note the domain restrictions, the denominator must not equal 0
02:47 And 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:

x3+1(x1)2=x+4 \frac{x^3+1}{(x-1)^2}=x+4

2

Step-by-step solution

To solve this equation, we follow these steps:

  • Step 1: Multiply both sides by (x1)2(x-1)^2 to eliminate the fraction.
  • Step 2: Expand and simplify both sides of the equation.
  • Step 3: Rearrange the equation to form a polynomial equal to zero.
  • Step 4: Solve the resulting polynomial using factorization or the quadratic formula.

Now, let's execute these steps:

Step 1: Multiply both sides by (x1)2(x-1)^2:
(x3+1)=(x+4)(x1)2(x^3 + 1) = (x + 4)(x - 1)^2

Step 2: Expand the right side:
(x+4)(x22x+1)=x(x22x+1)+4(x22x+1) (x + 4)(x^2 - 2x + 1) = x(x^2 - 2x + 1) + 4(x^2 - 2x + 1)

Calculating each part yields:
x(x22x+1)=x32x2+x x(x^2 - 2x + 1) = x^3 - 2x^2 + x
4(x22x+1)=4x28x+4 4(x^2 - 2x + 1) = 4x^2 - 8x + 4

Add these together:
x32x2+x+4x28x+4=x3+2x27x+4 x^3 - 2x^2 + x + 4x^2 - 8x + 4 = x^3 + 2x^2 - 7x + 4

Step 3: Combine terms and rearrange:
x3+1=x3+2x27x+4 x^3 + 1 = x^3 + 2x^2 - 7x + 4

Simplify by cancelling x3x^3 from both sides:
1=2x27x+4 1 = 2x^2 - 7x + 4

Move 1 to the right side:
0=2x27x+3 0 = 2x^2 - 7x + 3

Step 4: Solve the quadratic equation 2x27x+3=0 2x^2 - 7x + 3 = 0 .

Using the quadratic formula, x=b±b24ac2a x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} , where a=2 a = 2 , b=7 b = -7 , and c=3 c = 3 .

Calculate the discriminant:
b24ac=(7)2423=4924=25 b^2 - 4ac = (-7)^2 - 4 \cdot 2 \cdot 3 = 49 - 24 = 25

Now plug into the quadratic formula:
x=7±254 x = \frac{7 \pm \sqrt{25}}{4}

Simplify:
x=7±54 x = \frac{7 \pm 5}{4}

Two solutions arise:
x=124=3 x = \frac{12}{4} = 3 and x=24=12 x = \frac{2}{4} = \frac{1}{2}

Since x=1 x = 1 would make the denominator zero, it is not a valid solution for the original equation.

Therefore, the solution to the problem is x=3 x = 3 or x=12 x = \frac{1}{2} .

3

Final Answer

x=3,12 x=3,\frac{1}{2}

Key Points to Remember

Essential concepts to master this topic
  • Rule: Multiply both sides by the denominator to eliminate fractions
  • Technique: Expand (x+4)(x1)2 (x+4)(x-1)^2 to get x3+2x27x+4 x^3 + 2x^2 - 7x + 4
  • Check: Substitute x=3 x = 3 : 284=7 \frac{28}{4} = 7

Common Mistakes

Avoid these frequent errors
  • Forgetting to check for excluded values
    Don't skip checking if x=1 x = 1 makes the denominator zero = invalid solution! This critical step prevents division by zero errors. Always identify restricted values from the original denominator before solving.

Practice Quiz

Test your knowledge with interactive questions

\( (4b-3)(4b-3) \)

Rewrite the above expression as an exponential summation expression:

FAQ

Everything you need to know about this question

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

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Because x=1 x = 1 makes the denominator (x1)2=0 (x-1)^2 = 0 , which means division by zero. The original equation becomes undefined, so this value must be excluded from any solution set.

How do I expand (x+4)(x-1)² correctly?

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First expand (x1)2=x22x+1 (x-1)^2 = x^2 - 2x + 1 , then multiply by (x+4) (x+4) . Use distribution: x(x22x+1)+4(x22x+1) x(x^2-2x+1) + 4(x^2-2x+1) to get the final result.

What if I get a cubic equation instead of quadratic?

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Don't worry! In this problem, the x3 x^3 terms cancel out when you subtract, leaving a quadratic. Always combine like terms and simplify before assuming you need to solve a cubic.

Can I use factoring instead of the quadratic formula?

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Yes! For 2x27x+3=0 2x^2 - 7x + 3 = 0 , you can factor as (2x1)(x3)=0 (2x-1)(x-3) = 0 . This gives the same solutions: x=12 x = \frac{1}{2} and x=3 x = 3 .

How do I verify my solutions are correct?

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Substitute each solution back into the original equation. For x=3 x = 3 : 27+1(31)2=284=7 \frac{27+1}{(3-1)^2} = \frac{28}{4} = 7 and 3+4=7 3+4 = 7

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