If c is the longest side and a and b are the shorter sides, then the triangle is acute if [tex]c^2\ \textless \ a^2+b^2[/tex].
The longest side is 20, the shorter sides are x and x+4.
[tex]20^2 \ \textless \ x^2+(x+4)^2 \\
400\ \textless \ x^2+x^2+8x+16 \\
400\ \textless \ 2x^2+8x+16 \ \ \ \ \ \ \ \ \ \ \ \ \ |-400 \\
0\ \textless \ 2x^2+8x-384 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\div 2 \\
0\ \textless \ x^2+4x-192 \\
0\ \textless \ x^2+16x-12x-192 \\
0\ \textless \ x(x+16)-12(x+16) \\
0\ \textless \ (x-12)(x+16)[/tex]
The zeros are -16 and 12, the coefficient of x² is positive so the parabola opens upwards. The values greater than zero are in the interval [tex]x \in (-\infty, -16) \cup (12, \infty)[/tex].
The sides of the triangle must be positive numbers, so [tex]x\ \textgreater \ 12[/tex].
There's one number that satisfies this inequality: 14.
The answer is D.
