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

Question

(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

Video Solution

Solution Steps

00:00 Find X
00:03 When extracting a square root, both the numerator and denominator go under the root
00:17 Calculate the square root of 81 and square root of 64, and write
00:27 Multiply by the reciprocal to eliminate the fractions
00:40 Open parentheses properly, multiply by each factor
01:01 Calculate each fraction
01:10 Isolate X
01:25 This is the positive solution to the equation
01:35 Now let's calculate the negative solution to the equation
01:41 Remember - when extracting a square root there are two solutions: positive and negative
01:46 Multiply by the reciprocal to eliminate the fractions
01:50 Open parentheses properly, multiply by each factor
02:01 Isolate X
02:22 And this is the second (negative) solution to the equation
02:26 And this is the solution to the equation

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} .

Answer

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