Solve 12 131 1x x− + − =Give your answer in the form p q±2 where p and q are integers

1. Break Down the Problem

We start with the equation:

$12x^2 - 131x + 1 = 0$

We need to solve this quadratic equation for $x$ and express the solutions in the form $p \pm q\sqrt{2}$ where $p$ and $q$ are integers.

2. Relevant Concepts

To solve the quadratic equation $ax^2 + bx + c = 0$, we will use the quadratic formula:

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

In our case:

• $a = 12$
• $b = -131$
• $c = 1$

3. Analysis and Detail

Step 1: Calculate $b^2 - 4ac$

$b^2 - 4ac = (-131)^2 - 4(12)(1) = 17161 - 48 = 17113$

Step 2: Calculate the quadratic formula

Now we apply the quadratic formula:

$x = \frac{-(-131) \pm \sqrt{17113}}{2 \cdot 12} = \frac{131 \pm \sqrt{17113}}{24}$

4. Verify and Summarize

We need to approximate $\sqrt{17113}$ to check if it can be expressed as $q\sqrt{2}$.

$17113 = 2 \cdot 8556.5 \quad (\text{can be approximated but likely not an integer multiples})$

Therefore,

Final Answer

The solutions are:

$x = \frac{131 \pm \sqrt{17113}}{24}$

Since $\sqrt{17113}$ does not simplify to $q\sqrt{2}$ form with integer $p$ and $q$, we present it as:

$x = \frac{131}{24} \pm \frac{\sqrt{17113}}{24}$

However, in strict integer conditions $p$ and $q$ may vary based on approximation.