( 1 x + 1 2 ) 2 ( 1 x + 1 3 ) 2 = 81 64 \frac{(\frac{1}{x}+\frac{1}{2})^2}{(\frac{1}{x}+\frac{1}{3})^2}=\frac{81}{64} ( x 1 + 3 1 ) 2 ( x 1 + 2 1 ) 2 = 64 81
Find X
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 x x x .
Now, let's work through each step:
Step 1: Begin with the given equation:
( 1 x + 1 2 ) 2 ( 1 x + 1 3 ) 2 = 81 64 \frac{(\frac{1}{x} + \frac{1}{2})^2}{(\frac{1}{x} + \frac{1}{3})^2} = \frac{81}{64} ( x 1 + 3 1 ) 2 ( x 1 + 2 1 ) 2 = 64 81 .
Cross-multiply to eliminate fractions:
( 1 x + 1 2 ) 2 × 64 = ( 1 x + 1 3 ) 2 × 81 (\frac{1}{x} + \frac{1}{2})^2 \times 64 = (\frac{1}{x} + \frac{1}{3})^2 \times 81 ( x 1 + 2 1 ) 2 × 64 = ( x 1 + 3 1 ) 2 × 81 .
Step 2: Expand each squared term:
For ( 1 x + 1 2 ) 2 (\frac{1}{x} + \frac{1}{2})^2 ( x 1 + 2 1 ) 2 , use ( a + b ) 2 = a 2 + 2 a b + b 2 (a + b)^2 = a^2 + 2ab + b^2 ( a + b ) 2 = a 2 + 2 ab + b 2 :
( 1 x ) 2 + 2 ( 1 x ) ( 1 2 ) + ( 1 2 ) 2 = 1 x 2 + 1 x + 1 4 (\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} ( x 1 ) 2 + 2 ( x 1 ) ( 2 1 ) + ( 2 1 ) 2 = x 2 1 + x 1 + 4 1 .
Similarly, ( 1 x + 1 3 ) 2 = 1 x 2 + 2 3 x + 1 9 (\frac{1}{x} + \frac{1}{3})^2 = \frac{1}{x^2} + \frac{2}{3x} + \frac{1}{9} ( x 1 + 3 1 ) 2 = x 2 1 + 3 x 2 + 9 1 .
Step 3: Substitute these into the cross-multiplied equation:
64 ( 1 x 2 + 1 x + 1 4 ) = 81 ( 1 x 2 + 2 3 x + 1 9 ) 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) 64 ( x 2 1 + x 1 + 4 1 ) = 81 ( x 2 1 + 3 x 2 + 9 1 ) .
Step 4: Simplify and collect like terms:
64 ( 1 x 2 + 1 x + 1 4 ) = ( 64 1 x 2 + 64 1 x + 16 ) 64(\frac{1}{x^2} + \frac{1}{x} + \frac{1}{4}) = (64\frac{1}{x^2} + 64\frac{1}{x} + 16) 64 ( x 2 1 + x 1 + 4 1 ) = ( 64 x 2 1 + 64 x 1 + 16 ) ,
81 ( 1 x 2 + 2 3 x + 1 9 ) = ( 81 1 x 2 + 54 1 x + 9 ) 81(\frac{1}{x^2} + \frac{2}{3x} + \frac{1}{9}) = (81\frac{1}{x^2} + 54\frac{1}{x} + 9) 81 ( x 2 1 + 3 x 2 + 9 1 ) = ( 81 x 2 1 + 54 x 1 + 9 ) .
Equating terms gives:
64 1 x 2 + 64 1 x + 16 = 81 1 x 2 + 54 1 x + 9 64\frac{1}{x^2} + 64\frac{1}{x} + 16 = 81\frac{1}{x^2} + 54\frac{1}{x} + 9 64 x 2 1 + 64 x 1 + 16 = 81 x 2 1 + 54 x 1 + 9 .
Step 5: Solve the quadratic equation:
Combine like terms: − 17 1 x 2 + 10 1 x + 7 = 0 -17\frac{1}{x^2} + 10\frac{1}{x} + 7 = 0 − 17 x 2 1 + 10 x 1 + 7 = 0 .
Let y = 1 x y = \frac{1}{x} y = x 1 . Substitute to get: − 17 y 2 + 10 y + 7 = 0 -17y^2 + 10y + 7 = 0 − 17 y 2 + 10 y + 7 = 0 .
Multiply the entire equation by -1 to simplify: 17 y 2 − 10 y − 7 = 0 17y^2 - 10y - 7 = 0 17 y 2 − 10 y − 7 = 0 .
Using the quadratic formula y = − b ± b 2 − 4 a c 2 a y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} y = 2 a − b ± b 2 − 4 a c where a = 17 a=17 a = 17 , b = − 10 b=-10 b = − 10 , c = − 7 c=-7 c = − 7 :
y = 10 ± ( − 10 ) 2 − 4 × 17 × ( − 7 ) 2 × 17 y = \frac{10 \pm \sqrt{(-10)^2 - 4 \times 17 \times (-7)}}{2 \times 17} y = 2 × 17 10 ± ( − 10 ) 2 − 4 × 17 × ( − 7 )
y = 10 ± 100 + 476 34 y = \frac{10 \pm \sqrt{100 + 476}}{34} y = 34 10 ± 100 + 476
y = 10 ± 576 34 y = \frac{10 \pm \sqrt{576}}{34} y = 34 10 ± 576
y = 10 ± 24 34 y = \frac{10 \pm 24}{34} y = 34 10 ± 24
Which gives:
y = 34 34 = 1 y = \frac{34}{34} = 1 y = 34 34 = 1 or y = − 14 34 = − 7 17 y = -\frac{14}{34} = -\frac{7}{17} y = − 34 14 = − 17 7 .
Since y = 1 x y = \frac{1}{x} y = x 1 :
For y = 1 y=1 y = 1 , x = 1 y = 1 x = \frac{1}{y} = 1 x = y 1 = 1 .
For y = − 7 17 y=-\frac{7}{17} y = − 17 7 , x = 1 y = − 17 7 x = \frac{1}{y} = -\frac{17}{7} x = y 1 = − 7 17 .
Therefore, the solutions for x x x are x = 1 x = 1 x = 1 and x = − 17 7 x = -\frac{17}{7} x = − 7 17 .
Checking the correct answer choice, these correspond to the second choice.
Thus, the solution to the problem is x = 1 , − 17 7 x = 1, -\frac{17}{7} x = 1 , − 7 17 .
x = 1 , − 17 7 x=1,-\frac{17}{7} x = 1 , − 7 17