Question 456884
For the triangle to be acute, {{{a^2 + b^2 > c^2}}}, where c is the longest side of the triangle.  Note that it is given that the longest side has length 20.

Hence, we must have {{{x^2 + (x+4)^2 > 20^2}}}, or , after simplifying, {{{x^2 + 4x - 192 >0}}}, or (x+16)(x - 12) >0.  Since x must have positive values only, x > 12.

From the triangle inequality, we get the relation x +(x+4) > 20, or x > 8.  Also, since 20 is the longest side, we must have x + 4 < 20, or x < 16.  Hence from the initial conditions, we must have 8 < x < 16.

Intersect the preceding interval with the interval x > 12.

Therefore, for the triangle to be acute, we must have 12 < x < 16.