Question

1. if the sides of a triangle have the following lengths, find all the possible values for x.

\\( \overline { a b } = 4 x + 25, \overline { b c } = 3 x - 2, \overline { a c } = 9 x - 5 \\)

Explanation:

Step1: Apply triangle - side inequalities

For a triangle with sides \(a\), \(b\), and \(c\), the following inequalities must hold: \(a + b>c\), \(a + c>b\), and \(b + c>a\).
Let \(a = 4x + 25\), \(b=3x - 2\), \(c = 9x - 5\).
First, \(a + b>c\):
\((4x + 25)+(3x - 2)>9x - 5\)
\(4x+25 + 3x-2>9x - 5\)
\(7x + 23>9x - 5\)
\(23 + 5>9x-7x\)
\(28>2x\)
\(x < 14\).
Second, \(a + c>b\):
\((4x + 25)+(9x - 5)>3x - 2\)
\(4x+25 + 9x-5>3x - 2\)
\(13x + 20>3x - 2\)
\(13x-3x>-2 - 20\)
\(10x>-22\)
\(x>-\frac{11}{5}\).
Third, \(b + c>a\):
\((3x - 2)+(9x - 5)>4x + 25\)
\(3x-2 + 9x-5>4x + 25\)
\(12x-7>4x + 25\)
\(12x-4x>25 + 7\)
\(8x>32\)
\(x > 4\).
Also, consider the non - negativity of side lengths:
\(3x-2>0\Rightarrow x>\frac{2}{3}\), \(4x + 25>0\) (always true for real \(x\) since \(4x+25 = 0\Rightarrow x=-\frac{25}{4}\) and the coefficient of \(x\) is positive), \(9x - 5>0\Rightarrow x>\frac{5}{9}\).

Step2: Find the intersection of inequalities

We have the inequalities:
\(x < 14\), \(x>-\frac{11}{5}\), \(x > 4\), \(x>\frac{2}{3}\), \(x>\frac{5}{9}\).
The intersection of \(x>4\) and \(x < 14\) (since \(4>-\frac{11}{5}\), \(4>\frac{2}{3}\), \(4>\frac{5}{9}\)) gives the range of \(x\).

Answer:

\(4 < x < 14\)