A Pythagorean triple consists of three positive integers \(a\), \(b\), and \(c\) satisfying the equation \(a^2 + b^2 = c^2\), where \(c\) is the largest number.
In this case, we have the triple \((x, 12, 2x + 3)\). To solve for \(x\), we can assume that \(c = 2x + 3\) (the largest side), and we can write the equation:
\[ x^2 + 12^2 = (2x + 3)^2 \]
Calculating \(12^2\):
\[ 12^2 = 144 \]
Now we have:
\[ x^2 + 144 = (2x + 3)^2 \]
Expanding the right side:
\[ (2x + 3)^2 = 4x^2 + 12x + 9 \]
Substituting this back into the equation gives:
\[ x^2 + 144 = 4x^2 + 12x + 9 \]
Now, rearranging the equation to bring all the terms to one side:
\[ x^2 - 4x^2 - 12x + 144 - 9 = 0 \]
This simplifies to:
\[ -3x^2 - 12x + 135 = 0 \]
Multiplying the entire equation by -1 to simplify:
\[ 3x^2 + 12x - 135 = 0 \]
Now we can use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 3\), \(b = 12\), and \(c = -135\):
Calculating the discriminant:
\[ b^2 - 4ac = 12^2 - 4 \cdot 3 \cdot (-135) = 144 + 1620 = 1764 \]
Now, calculating \(x\):
\[ x = \frac{-12 \pm \sqrt{1764}}{2 \cdot 3} \] \[ \sqrt{1764} = 42 \] \[ x = \frac{-12 \pm 42}{6} \]
This gives us two possible values for \(x\):
- \(x = \frac{30}{6} = 5\)
- \(x = \frac{-54}{6} = -9\) (which we discard since \(x\) must be positive).
Thus, the value of \(x\) is:
\(\boxed{5}\)
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