Find the Difference!

Let $a=\sqrt{11x^{2}-21x+31}$ and $b=\sqrt{11x^{2}-21x-9}$

Let $I=\sqrt{11x^{2}-21x+31}+\sqrt{11x^{2}-21x-9}$ and $J=\sqrt{11x^{2}-21x+31}-\sqrt{11x^{2}-21x-9}$

Therefore, $I=a+b$ and $J=a-b$

By the difference of two squares, we have: $IJ=(a+b)(a-b)=a^{2}-b^{2}$ $IJ=(11^{2}-21x+31)-(11x^{2}-21x-9)$ $IJ=40$

$I=8$ $J=\frac{40}{8}=\boxed{5}$

Let's define $a^2 = 11x^2 - 21x + 31.$

$\sqrt{a^2} + \sqrt{a^2 - 40} = 8$

$8-a = \sqrt{a^2 - 40}$

$a^2 - 16a + 64 = a^2 - 40 => a = \boxed{6.5}$

The required is: $a - \sqrt{a-40} = 6.5 - \sqrt{1.75} = \boxed{5}$

Let's suppose that a=V(11x^2 -21x +31) and b=V(11x62-21x-9) So we have : Va+Vb=8, then 1/8 = 1/(Va+Vb), by rationalizing, we get 8(Va-Vb)= (a-b), therefore 8(Va-Vb)=40 So, Va-Vb=40/8=5

We must know if (11x^2 - 21x + 31)^1/2 = 13/2, so (11x^2 - 21x - 9)^1/2 = (11x^2 - 21x + 31 - 40 )^1/2 = 3/2. So, (11x^2 - 21x + 31)^1/2 - (11x^2 - 21x - 9)^1/2 = 13/2 - 3/2 = 10/2 = 5.

Let $11x^2-21x-9=p$ then $11x^2-21x+31=p+40$ . Now we can modify the first equation to the form $\sqrt {p+40}+\sqrt {p}=8$ . And by the same way, the desired form can change to be $\sqrt {p+40}-\sqrt {p}$ .

Note that $(\sqrt {p+40}+\sqrt {p})(\sqrt {p+40}-\sqrt {p})=(p+40)-p=40$ . By this, we have a new equation now

$(\sqrt {p+40}+\sqrt {p})(\sqrt {p+40}-\sqrt {p})=40$

$8(\sqrt {p+40}-\sqrt {p})=40$

$\sqrt {p+40}-\sqrt {p}=\boxed {5}$