Answer :
Sure, let's solve the given problem step-by-step.
First, we need to solve the equation:
[tex]\[ x^2 + \frac{1}{25 x^2} = 8 \frac{3}{5} \][/tex]
1. Convert the mixed fraction to an improper fraction:
[tex]\[ 8 \frac{3}{5} = \frac{8 \times 5 + 3}{5} = \frac{40 + 3}{5} = \frac{43}{5} \][/tex]
So, the equation becomes:
[tex]\[ x^2 + \frac{1}{25 x^2} = \frac{43}{5} \][/tex]
2. Multiply both sides by 25x² to clear the fraction:
[tex]\[ 25 x^2 \left( x^2 + \frac{1}{25 x^2} \right) = 25 x^2 \cdot \frac{43}{5} \][/tex]
[tex]\[ 25 x^4 + 1 = \frac{43}{5} \cdot 25 x^2 \][/tex]
[tex]\[ 25 x^4 + 1 = 215 x^2 \][/tex]
3. Rearrange the equation to standard quadratic form:
[tex]\[ 25 x^4 - 215 x^2 + 1 = 0 \][/tex]
Let [tex]\( y = x^2 \)[/tex]. The equation becomes:
[tex]\[ 25 y^2 - 215 y + 1 = 0 \][/tex]
This is a quadratic equation in [tex]\( y \)[/tex]. We can solve it using the quadratic formula [tex]\( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 25 \)[/tex], [tex]\( b = -215 \)[/tex], and [tex]\( c = 1 \)[/tex]:
[tex]\[ y = \frac{-(-215) \pm \sqrt{(-215)^2 - 4 \cdot 25 \cdot 1}}{2 \cdot 25} \][/tex]
[tex]\[ y = \frac{215 \pm \sqrt{46225 - 100}}{50} \][/tex]
[tex]\[ y = \frac{215 \pm \sqrt{46125}}{50} \][/tex]
[tex]\[ y = \frac{215 \pm 214.85}{50} \][/tex]
So, we get two solutions for [tex]\( y \)[/tex]:
[tex]\[ y_1 = \frac{215 + 214.85}{50} = \frac{429.85}{50} = 8.597 \][/tex]
[tex]\[ y_2 = \frac{215 - 214.85}{50} = \frac{0.15}{50} = 0.003 \][/tex]
Recall that [tex]\( y = x^2 \)[/tex], so we have:
[tex]\( x^2 = 8.597 \)[/tex] [tex]\(\text{or}\)[/tex] [tex]\( x^2 = 0.003 \)[/tex].
Thus, we find [tex]\( x \)[/tex] as:
[tex]\[ x_1 = \sqrt{8.597} \approx 2.931 \][/tex]
[tex]\[ x_2 = -\sqrt{8.597} \approx -2.931 \][/tex]
[tex]\[ x_3 = \sqrt{0.003} \approx 0.068 \][/tex]
[tex]\[ x_4 = -\sqrt{0.003} \approx -0.068 \][/tex]
4. Calculate the value of [tex]\( x^3 + \frac{1}{125 x^3} \)[/tex] for each solution:
For [tex]\( x_1 \approx 2.931 \)[/tex]:
[tex]\[ x_1^3 + \frac{1}{125 x_1^3} \approx 25.2 \][/tex]
For [tex]\( x_2 \approx -2.931 \)[/tex]:
[tex]\[ x_2^3 + \frac{1}{125 x_2^3} \approx -25.2 \][/tex]
For [tex]\( x_3 \approx 0.068 \)[/tex]:
[tex]\[ x_3^3 + \frac{1}{125 x_3^3} \approx 25.2 \][/tex]
For [tex]\( x_4 \approx -0.068 \)[/tex]:
[tex]\[ x_4^3 + \frac{1}{125 x_4^3} \approx -25.2 \][/tex]
Thus, the values of [tex]\( x^3 + \frac{1}{125 x^3} \)[/tex] for the given solutions are approximately:
[tex]\[ 25.2, -25.2, 25.2, -25.2 \][/tex]
First, we need to solve the equation:
[tex]\[ x^2 + \frac{1}{25 x^2} = 8 \frac{3}{5} \][/tex]
1. Convert the mixed fraction to an improper fraction:
[tex]\[ 8 \frac{3}{5} = \frac{8 \times 5 + 3}{5} = \frac{40 + 3}{5} = \frac{43}{5} \][/tex]
So, the equation becomes:
[tex]\[ x^2 + \frac{1}{25 x^2} = \frac{43}{5} \][/tex]
2. Multiply both sides by 25x² to clear the fraction:
[tex]\[ 25 x^2 \left( x^2 + \frac{1}{25 x^2} \right) = 25 x^2 \cdot \frac{43}{5} \][/tex]
[tex]\[ 25 x^4 + 1 = \frac{43}{5} \cdot 25 x^2 \][/tex]
[tex]\[ 25 x^4 + 1 = 215 x^2 \][/tex]
3. Rearrange the equation to standard quadratic form:
[tex]\[ 25 x^4 - 215 x^2 + 1 = 0 \][/tex]
Let [tex]\( y = x^2 \)[/tex]. The equation becomes:
[tex]\[ 25 y^2 - 215 y + 1 = 0 \][/tex]
This is a quadratic equation in [tex]\( y \)[/tex]. We can solve it using the quadratic formula [tex]\( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 25 \)[/tex], [tex]\( b = -215 \)[/tex], and [tex]\( c = 1 \)[/tex]:
[tex]\[ y = \frac{-(-215) \pm \sqrt{(-215)^2 - 4 \cdot 25 \cdot 1}}{2 \cdot 25} \][/tex]
[tex]\[ y = \frac{215 \pm \sqrt{46225 - 100}}{50} \][/tex]
[tex]\[ y = \frac{215 \pm \sqrt{46125}}{50} \][/tex]
[tex]\[ y = \frac{215 \pm 214.85}{50} \][/tex]
So, we get two solutions for [tex]\( y \)[/tex]:
[tex]\[ y_1 = \frac{215 + 214.85}{50} = \frac{429.85}{50} = 8.597 \][/tex]
[tex]\[ y_2 = \frac{215 - 214.85}{50} = \frac{0.15}{50} = 0.003 \][/tex]
Recall that [tex]\( y = x^2 \)[/tex], so we have:
[tex]\( x^2 = 8.597 \)[/tex] [tex]\(\text{or}\)[/tex] [tex]\( x^2 = 0.003 \)[/tex].
Thus, we find [tex]\( x \)[/tex] as:
[tex]\[ x_1 = \sqrt{8.597} \approx 2.931 \][/tex]
[tex]\[ x_2 = -\sqrt{8.597} \approx -2.931 \][/tex]
[tex]\[ x_3 = \sqrt{0.003} \approx 0.068 \][/tex]
[tex]\[ x_4 = -\sqrt{0.003} \approx -0.068 \][/tex]
4. Calculate the value of [tex]\( x^3 + \frac{1}{125 x^3} \)[/tex] for each solution:
For [tex]\( x_1 \approx 2.931 \)[/tex]:
[tex]\[ x_1^3 + \frac{1}{125 x_1^3} \approx 25.2 \][/tex]
For [tex]\( x_2 \approx -2.931 \)[/tex]:
[tex]\[ x_2^3 + \frac{1}{125 x_2^3} \approx -25.2 \][/tex]
For [tex]\( x_3 \approx 0.068 \)[/tex]:
[tex]\[ x_3^3 + \frac{1}{125 x_3^3} \approx 25.2 \][/tex]
For [tex]\( x_4 \approx -0.068 \)[/tex]:
[tex]\[ x_4^3 + \frac{1}{125 x_4^3} \approx -25.2 \][/tex]
Thus, the values of [tex]\( x^3 + \frac{1}{125 x^3} \)[/tex] for the given solutions are approximately:
[tex]\[ 25.2, -25.2, 25.2, -25.2 \][/tex]
