Describe the possible values of x in the figure shown

Question

Sides of the triangle are 2x+1, 2x+31, x+16

Answer 1

I am assuming that the figure mentioned is a right-angles triangle.

If so, then the hypotenuse is the longest side.

i.e. hypotenuse = 2x + 31, since 2x + 31 is greater than either 2x + 1 or x + 16.

Using pythagoras' theorem,

Hypotenuse squared = sum of squares of other two sides.

i.e. (2x + 31)^2 = (2x + 1)^2 + (x + 16)^2

4x^2 + 124x + 961 = 4x^2 + 4x + 1 + x^2 + 32x + 256

x^2 + (4 + 32 - 124)x + (1 + 256 - 961) = 0

x^2 - 88x - 704 = 0

Using the quadratic formula,

x = 44 +/- sqrt(165)

x = 44 +/-  sqrt(165) = 44 +/-  51.3809

x= 95.38, x = -7.38

If x = -7.38, then one side of the triangle will be 2x + 1 = -14.76 + 1 = negative.

We cannot have a negative length, so the negative solution for x should be ignored

 

Answer 2

In the absence of a diagram, we can use the cosine rule to figure out the range for x:

(2x+31)^2=(2x+1)^2+(x+16)^2-2(2x+1)(x+16)cosA where A is an angle.

(2x+31)^2-(2x+1)^2=(x+16)(x+16-(4x+2)cosA);

(2x+31-2x-1)(2x+31+2x+1)=(x+16)(x+16-(4x+2)cosA);

30(4x+32)=(x+16)(x+16-(4x+2)cosA).

cosA must lie between -1 and 1: -1<cosA<1; or -1<(x^2-88x-704)/(2(x+16)(2x+1))<1;

-(4x^2+66x+32)<x^2-88x-704<4x^2+66x+32 (this cross-multiplication is only valid if the denominator is positive; otherwise the inequality has to be reversed; assume for now that it is positive);

5x^2-22x-672>0, (5x+48)(x-14)>0.

So, x>14 or x<-48/5; note that x>14 makes 2(x+16)(2x+1) positive, but x<-48/5 makes it negative, so the inequality is reversed and x>-48/5.

Alternatively:

3x^2+154x+736>0, (3x+16)(x+46)>0. This has solutions x<-46 and x>-16/3; the first makes 2(x+16)(2x+1) positive and the second makes it negative, so the inequality must be reversed, and x<-16/3.

We can see that (x^2-88x-704)/(2(x+16)(2x+1)) lies between the required limits for cosA when -48/5<x<-16/3 and x>14, and x<-46, which can be rejected because it would result in all the sides of the triangle being negative.

For the remaining ranges we have the following values for (x+16, 2x+1, 2x+31) on (-48/5,-16/3,14):

(6.4,-18.2,11.8); (10.67,-9.67,20.33); (30,29,59). The first two of these can be rejected because the sides of the triangle must be positive, leaving x>14 as the only solution.