Question 1191818
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Finding the maximum possible value of x is easy.  It is given that the longest side is 45, so<br>
{{{x+7<45}}}
{{{x<38}}}<br>
To determine the smallest possible values of x, note that the triangle is a right triangle if<br>
{{{(x-4)^2+(x+7)^2=45^2}}}<br>
So the triangle is acute if<br>
{{{(x-4)^2+(x+7)^2>45^2}}}<br>
{{{(x-4)^2+(x+7)^2=45^2}}}
{{{x^2-8x+16+x^2+14x+49=2025}}}
{{{2x^2+6x+65=2025}}}
{{{2x^2+6x-1960=0}}}
{{{x^2+3x-980=0}}}<br>
Use the quadratic formula to find<br>
{{{x=(-3+sqrt(3929))/2}}} = 29.805 to 3 decimal places<br>
So the triangle with longest side 45 and the other two sides x-4 and x+7 is acute for any value greater than (approximately) 29.805 and less than 38.<br>
ANSWER: 29.805 < x < 38<br>