Solve for X in the Fractional Power Equation: ((1/x)+(1/2))² / ((1/x)+(1/3))² = 81/64

Rational Equations with Cross Multiplication

(1x+12)2(1x+13)2=8164 \frac{(\frac{1}{x}+\frac{1}{2})^2}{(\frac{1}{x}+\frac{1}{3})^2}=\frac{81}{64}

Find X

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations
00:11 Find X. Let's see how we can solve for X step-by-step.
00:16 When you find a square root, both the top and bottom number s, go under the root.
00:28 Now, find the square root of 81 and the square root of 64. Let's w rite them out.
00:38 Next, to get rid of the fractions, multiply by the reciprocal.
00:51 Be sure to open the parentheses correctly. Then multiply by each f actor.
01:12 Now, simply calculate each fraction.
01:21 Let's isolate X. This will help find our answer.
01:36 We have found the positive solution to the equation.
01:46 Now, let's work on the negative solution of the equation.
01:52 Remember, when finding a square root, there are two solutio ns: positive and negative.
01:58 Multiply by the reciprocal again to clear the fractions.
02:03 Open the parentheses correctly and multiply by each factor again .
02:12 Isolate X once more to find our other solution.
02:33 And this is the second, or negative, solution to the equation .
02:38 Great job! Now you know the solutions to the equation.

Step-by-step written solution

Follow each step carefully to understand the complete solution
1

Understand the problem

(1x+12)2(1x+13)2=8164 \frac{(\frac{1}{x}+\frac{1}{2})^2}{(\frac{1}{x}+\frac{1}{3})^2}=\frac{81}{64}

Find X

2

Step-by-step solution

To solve this problem, we'll follow these steps:

  • Step 1: Cross-multiply the given equation to eliminate fractions.
  • Step 2: Expand the squared terms on either side of the equation.
  • Step 3: Rearrange terms to form a quadratic equation.
  • Step 4: Solve the quadratic equation to find possible values for xx.

Now, let's work through each step:

Step 1: Begin with the given equation:
(1x+12)2(1x+13)2=8164\frac{(\frac{1}{x} + \frac{1}{2})^2}{(\frac{1}{x} + \frac{1}{3})^2} = \frac{81}{64}. Cross-multiply to eliminate fractions:
(1x+12)2×64=(1x+13)2×81(\frac{1}{x} + \frac{1}{2})^2 \times 64 = (\frac{1}{x} + \frac{1}{3})^2 \times 81.

Step 2: Expand each squared term:
For (1x+12)2(\frac{1}{x} + \frac{1}{2})^2, use (a+b)2=a2+2ab+b2(a + b)^2 = a^2 + 2ab + b^2:
(1x)2+2(1x)(12)+(12)2=1x2+1x+14(\frac{1}{x})^2 + 2(\frac{1}{x})(\frac{1}{2}) + (\frac{1}{2})^2 = \frac{1}{x^2} + \frac{1}{x} + \frac{1}{4}.
Similarly, (1x+13)2=1x2+23x+19(\frac{1}{x} + \frac{1}{3})^2 = \frac{1}{x^2} + \frac{2}{3x} + \frac{1}{9}.

Step 3: Substitute these into the cross-multiplied equation:
64(1x2+1x+14)=81(1x2+23x+19)64\left(\frac{1}{x^2} + \frac{1}{x} + \frac{1}{4}\right) = 81\left(\frac{1}{x^2} + \frac{2}{3x} + \frac{1}{9}\right).

Step 4: Simplify and collect like terms:
64(1x2+1x+14)=(641x2+641x+16)64(\frac{1}{x^2} + \frac{1}{x} + \frac{1}{4}) = (64\frac{1}{x^2} + 64\frac{1}{x} + 16),
81(1x2+23x+19)=(811x2+541x+9)81(\frac{1}{x^2} + \frac{2}{3x} + \frac{1}{9}) = (81\frac{1}{x^2} + 54\frac{1}{x} + 9).

Equating terms gives:
641x2+641x+16=811x2+541x+964\frac{1}{x^2} + 64\frac{1}{x} + 16 = 81\frac{1}{x^2} + 54\frac{1}{x} + 9.

Step 5: Solve the quadratic equation:
Combine like terms: 171x2+101x+7=0-17\frac{1}{x^2} + 10\frac{1}{x} + 7 = 0.
Let y=1xy = \frac{1}{x}. Substitute to get: 17y2+10y+7=0-17y^2 + 10y + 7 = 0.
Multiply the entire equation by -1 to simplify: 17y210y7=017y^2 - 10y - 7 = 0.

Using the quadratic formula y=b±b24ac2ay = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} where a=17a=17, b=10b=-10, c=7c=-7:
y=10±(10)24×17×(7)2×17 y = \frac{10 \pm \sqrt{(-10)^2 - 4 \times 17 \times (-7)}}{2 \times 17}
y=10±100+47634 y = \frac{10 \pm \sqrt{100 + 476}}{34}
y=10±57634 y = \frac{10 \pm \sqrt{576}}{34}
y=10±2434 y = \frac{10 \pm 24}{34}
Which gives:
y=3434=1 y = \frac{34}{34} = 1 or y=1434=717 y = -\frac{14}{34} = -\frac{7}{17} .

Since y=1xy = \frac{1}{x}:
For y=1y=1, x=1y=1x = \frac{1}{y} = 1.
For y=717y=-\frac{7}{17}, x=1y=177x = \frac{1}{y} = -\frac{17}{7}.

Therefore, the solutions for xx are x=1x = 1 and x=177x = -\frac{17}{7}.

Checking the correct answer choice, these correspond to the second choice.

Thus, the solution to the problem is x=1,177 x = 1, -\frac{17}{7} .

3

Final Answer

x=1,177 x=1,-\frac{17}{7}

Key Points to Remember

Essential concepts to master this topic
  • Rule: Cross multiply first to eliminate complex fraction ratios
  • Technique: Expand (1x+12)2 (\frac{1}{x} + \frac{1}{2})^2 using (a+b)2=a2+2ab+b2 (a+b)^2 = a^2 + 2ab + b^2
  • Check: Substitute solutions back into original equation to verify both sides equal 8164 \frac{81}{64}

Common Mistakes

Avoid these frequent errors
  • Forgetting to square both terms when expanding
    Don't expand (1x+12)2 (\frac{1}{x} + \frac{1}{2})^2 as just 1x2+14 \frac{1}{x^2} + \frac{1}{4} = missing the cross term! This loses the 2(1x)(12)=1x 2(\frac{1}{x})(\frac{1}{2}) = \frac{1}{x} term and gives wrong coefficients. Always use the complete formula (a+b)2=a2+2ab+b2 (a+b)^2 = a^2 + 2ab + b^2 for every term.

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 do we cross multiply instead of finding a common denominator?

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Cross multiplication is faster when you have one fraction equals another fraction. Since we have A2B2=8164 \frac{A^2}{B^2} = \frac{81}{64} , cross multiplying gives us 64A2=81B2 64A^2 = 81B^2 directly!

What does the substitution y = 1/x actually do?

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This substitution trick transforms our complex rational equation into a simple quadratic! Instead of dealing with 1x2 \frac{1}{x^2} and 1x \frac{1}{x} terms, we get y2 y^2 and y y which are much easier to solve.

How do I know which answer choices to eliminate?

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Check the signs carefully! We found y=1 y = 1 and y=717 y = -\frac{7}{17} . Since x=1y x = \frac{1}{y} , we get x=1 x = 1 and x=177 x = -\frac{17}{7} (note the sign flip!).

Why do we get two solutions for x?

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Because our substitution led to a quadratic equation 17y210y7=0 17y^2 - 10y - 7 = 0 ! Quadratic equations typically have two solutions, and each y-value gives us a corresponding x-value.

What if I get confused expanding the squared terms?

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Break it down step by step! For (1x+12)2 (\frac{1}{x} + \frac{1}{2})^2 :

  • First term squared: (1x)2=1x2 (\frac{1}{x})^2 = \frac{1}{x^2}
  • Cross term: 21x12=1x 2 \cdot \frac{1}{x} \cdot \frac{1}{2} = \frac{1}{x}
  • Last term squared: (12)2=14 (\frac{1}{2})^2 = \frac{1}{4}

How can I verify my final answers are correct?

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Always substitute back! For x=1 x = 1 : (1+12)2(1+13)2=(32)2(43)2=94169=8164 \frac{(1 + \frac{1}{2})^2}{(1 + \frac{1}{3})^2} = \frac{(\frac{3}{2})^2}{(\frac{4}{3})^2} = \frac{\frac{9}{4}}{\frac{16}{9}} = \frac{81}{64}

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