Math Challenge | Grades 9-10 | Rational Equation Problem with Long Solution [in Filipino] - Free Printable

We are tasked with solving the equation:

\[
2 \left( x + \frac{1}{x} \right)^2 = 10 - x - \frac{1}{x}
\]

Step 1: Simplify the equation

Let us introduce a substitution to simplify the expression. Let:

\[
y = x + \frac{1}{x}
\]

Then the equation becomes:

\[
2y^2 = 10 - y
\]

Step 2: Rearrange the equation

Rearrange the equation to form a standard quadratic equation:

\[
2y^2 + y - 10 = 0
\]

Step 3: Solve the quadratic equation

We solve the quadratic equation \(2y^2 + y - 10 = 0\) using the quadratic formula:

\[
y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Here, \(a = 2\), \(b = 1\), and \(c = -10\). Substituting these values into the formula:

\[
y = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 2 \cdot (-10)}}{2 \cdot 2}
\]

Simplify inside the square root:

\[
y = \frac{-1 \pm \sqrt{1 + 80}}{4}
\]

\[
y = \frac{-1 \pm \sqrt{81}}{4}
\]

\[
y = \frac{-1 \pm 9}{4}
\]

This gives us two solutions for \(y\):

\[
y = \frac{-1 + 9}{4} = \frac{8}{4} = 2
\]

\[
y = \frac{-1 - 9}{4} = \frac{-10}{4} = -\frac{5}{2}
\]

Thus, the possible values for \(y\) are:

\[
y = 2 \quad \text{or} \quad y = -\frac{5}{2}
\]

Step 4: Solve for \(x\) using \(y = x + \frac{1}{x}\)

#### Case 1: \(y = 2\)
We have:

\[
x + \frac{1}{x} = 2
\]

Multiply through by \(x\) (assuming \(x \neq 0\)):

\[
x^2 + 1 = 2x
\]

Rearrange into standard quadratic form:

\[
x^2 - 2x + 1 = 0
\]

Factor the quadratic:

\[
(x - 1)^2 = 0
\]

Thus:

\[
x = 1
\]

#### Case 2: \(y = -\frac{5}{2}\)
We have:

\[
x + \frac{1}{x} = -\frac{5}{2}
\]

Multiply through by \(x\) (assuming \(x \neq 0\)):

\[
x^2 + 1 = -\frac{5}{2}x
\]

Rearrange into standard quadratic form:

\[
x^2 + \frac{5}{2}x + 1 = 0
\]

To eliminate the fraction, multiply through by 2:

\[
2x^2 + 5x + 2 = 0
\]

Solve this quadratic equation using the quadratic formula, where \(a = 2\), \(b = 5\), and \(c = 2\):

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

\[
x = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 2 \cdot 2}}{2 \cdot 2}
\]

\[
x = \frac{-5 \pm \sqrt{25 - 16}}{4}
\]

\[
x = \frac{-5 \pm \sqrt{9}}{4}
\]

\[
x = \frac{-5 \pm 3}{4}
\]

This gives us two solutions:

\[
x = \frac{-5 + 3}{4} = \frac{-2}{4} = -\frac{1}{2}
\]

\[
x = \frac{-5 - 3}{4} = \frac{-8}{4} = -2
\]

Step 5: Verify the solutions

We need to verify that each solution satisfies the original equation.

#### Verify \(x = 1\):
Substitute \(x = 1\) into the original equation:

\[
2 \left( 1 + \frac{1}{1} \right)^2 = 10 - 1 - \frac{1}{1}
\]

\[
2 (1 + 1)^2 = 10 - 1 - 1
\]

\[
2 (2)^2 = 10 - 2
\]

\[
2 \cdot 4 = 8
\]

\[
8 = 8 \quad \text{(True)}
\]

#### Verify \(x = -\frac{1}{2}\):
Substitute \(x = -\frac{1}{2}\) into the original equation:

\[
2 \left( -\frac{1}{2} + \frac{1}{-\frac{1}{2}} \right)^2 = 10 - \left( -\frac{1}{2} \right) - \frac{1}{-\frac{1}{2}}
\]

\[
2 \left( -\frac{1}{2} - 2 \right)^2 = 10 + \frac{1}{2} + 2
\]

\[
2 \left( -\frac{5}{2} \right)^2 = 10 + \frac{1}{2} + 2
\]

\[
2 \left( \frac{25}{4} \right) = 12 + \frac{1}{2}
\]

\[
\frac{50}{4} = \frac{24}{2} + \frac{1}{2}
\]

\[
\frac{25}{2} = \frac{25}{2} \quad \text{(True)}
\]

#### Verify \(x = -2\):
Substitute \(x = -2\) into the original equation:

\[
2 \left( -2 + \frac{1}{-2} \right)^2 = 10 - (-2) - \frac{1}{-2}
\]

\[
2 \left( -2 - \frac{1}{2} \right)^2 = 10 + 2 + \frac{1}{2}
\]

\[
2 \left( -\frac{5}{2} \right)^2 = 12 + \frac{1}{2}
\]

\[
2 \left( \frac{25}{4} \right) = \frac{24}{2} + \frac{1}{2}
\]

\[
\frac{50}{4} = \frac{25}{2}
\]

\[
\frac{25}{2} = \frac{25}{2} \quad \text{(True)}
\]

Final Answer:

The solutions to the equation are:

\[
\boxed{1, -\frac{1}{2}, -2}
\]