If (135)x + (144) x = (323)x, what is the value of x?

The given equation is:

\[
135_x + 144_x = 323_x
\]

Where the subscript \( x \) represents that the numbers are in base \( x \).

To solve for \( x \), let's first convert the numbers from base \( x \) to base 10:

1. **Convert \( 135_x \) to base 10:**
   \[
   135_x = 1 \cdot x^2 + 3 \cdot x^1 + 5 \cdot x^0 = x^2 + 3x + 5
   \]

2. **Convert \( 144_x \) to base 10:**
   \[
   144_x = 1 \cdot x^2 + 4 \cdot x^1 + 4 \cdot x^0 = x^2 + 4x + 4
   \]

3. **Convert \( 323_x \) to base 10:**
   \[
   323_x = 3 \cdot x^2 + 2 \cdot x^1 + 3 \cdot x^0 = 3x^2 + 2x + 3
   \]

Now, substitute these into the equation:

\[
(x^2 + 3x + 5) + (x^2 + 4x + 4) = 3x^2 + 2x + 3
\]

Simplify the left-hand side:

\[
2x^2 + 7x + 9 = 3x^2 + 2x + 3
\]

Now, bring all terms to one side:

\[
0 = 3x^2 + 2x + 3 - 2x^2 - 7x - 9
\]

Simplify further:

\[
0 = x^2 - 5x - 6
\]

This is a quadratic equation, which can be factored as:

\[
0 = (x - 6)(x + 1)
\]

So, the possible values for \( x \) are:

\[
x = 6 \quad \text{or} \quad x = -1
\]

Since \( x \) must be a positive integer (since it's a base), we discard \( x = -1 \) and are left with:

\[
x = 6
\]

**Final Answer: \( x = 6 \)**

Would you like any more details or explanations?

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Here are some related questions:

1. How do you convert a number from base \( x \) to base 10?
2. What is the general process for solving equations involving different bases?
3. Can a base \( x \) be negative or a non-integer?
4. What would happen if the quadratic equation had no real solutions?
5. How do you factorize quadratic equations?
6. How do different bases affect the difficulty of solving an equation?
7. What is the significance of a base in mathematics?
8. How would the equation change if one of the numbers were in a different base?

**Tip:** Always check if your quadratic equation can be factored easily before using the quadratic formula.

Learn how to solve the equation 135_x + 144_x = 323_x in base x. Understand base conversions and quadratic equations with a step-by-step solution. Suitable for high school students studying algebra and number systems.

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