The longest side of the triangle is 20, another side length is 'x + 4'
Since the triangle is acute, we know that the value of 'x' will have to be <u>greater than its right triangle value</u> because whatever 'x' equals will result in a <em>right </em>triangle, not an acute one
We must solve for the <em>right </em>triangle first in order to determine the acute triangle
Solving for 'x' in the right triangle:
a^2 + b^2 = c^2
(x+4)^2 + x^2 = 20^2
x^2 + x^2 + 8x + 16 = 400
2x^2 + 8x - 384 = 0
simplify x^2 + 4x - 192 = 0
factor (x + 16) (x - 12) = 0
solve for each 'x' x + 16 = 0 & x - 12 = 0
x = -16 & x = 12
check the value of 'x' cannot be negative, so it is not -16. this means the value of 'x' is 12
the value of 12 for 'x' makes a <em>right triangle, </em>so that is not the answer. this means the the value of 'x' needs to be <em>greater</em> than 12
<u>Checking your answer:</u>
original equation (x + 4)^2 + x^2= 20^2
with answer (12 + 4)^2 + (12)^2= 20^2
(16)^2 + (12)^2 = 20^2
256 + 144 = 400
400 = 400 this is true, so the answer of 'x = 12' is <em>correct </em>for making it into a <em>right triangle</em>
<em></em>
Solving for 'x' in the acute triangle:
x = 12 makes a right triangle
x > 12 minimum for an acute triangle
what is the maximum value for 'x' in the acute triangle? take the two other side lengths --> (x + 4) < 20 (longest side, so it will be the greater number)
simplify x < 16
x < 16 maximum for an acute triangle
Conclusion:
if 'x = 12' makes a right triangle, then 'x > 12' and 'x < 16' are the values that will make it acute
