If a polynomial ax² + bx + c is a perfect square if b² = 4ac

For the polynomial x² - 11x + m

a = 1
b = -11
c = m

Since the polynomial x² - 11x + m is a perfect square, b² = 4ac

(-11)² = 4 x 1 x m

121 = 4m

Divide both sides by 4

m = $\frac{121}{4}$

p² + q² + r² = 50

p = 5

$\sqrt{q}$ = 2

Square both sides to get rid of the square root.

q = 2² = 4

Substitute the known values into the equation.

p² + q² + r² = 50

5² + 4² + r² = 50

25 + 16 + r² = 50

41 + r² = 50

r² = 50 - 41

r² = 9

Take the square root of both sides to get rid of the square

r = $\sqrt{9}$ = 3

$\frac{2}{\mathrm{(x\; -\; 3)}}$ - $\frac{3}{\mathrm{(x\; -\; 2)}}$ = $\frac{2}{\mathrm{(x\; -\; 3)}}$ - $\frac{3}{\mathrm{(x\; -\; 2)}}$

$\frac{2}{\mathrm{(x\; -\; 3)}}$ - $\frac{3}{\mathrm{(x\; -\; 2)}}$ = $\frac{\mathrm{2\; x\; (x\; -\; 2)}-\mathrm{3\; x\; (x\; -\; 3)}}{\mathrm{(x\; -\; 3)((x\; -\; 2)}}$

$\frac{2}{\mathrm{(x\; -\; 3)}}$ - $\frac{3}{\mathrm{(x\; -\; 2)}}$ = $\frac{\mathrm{2x\; -\; 4\; -\; 3x\; +\; 9}}{\mathrm{(x\; -\; 3)((x\; -\; 2)}}$

$\frac{2}{\mathrm{(x\; -\; 3)}}$ - $\frac{3}{\mathrm{(x\; -\; 2)}}$ = $\frac{\mathrm{2x\; -\; 3x\; +\; 9\; -\; 4}}{\mathrm{(x\; -\; 3)((x\; -\; 2)}}$

$\frac{2}{\mathrm{(x\; -\; 3)}}$ - $\frac{3}{\mathrm{(x\; -\; 2)}}$ = $\frac{\mathrm{-x\; +\; 5}}{\mathrm{(x\; -\; 3)((x\; -\; 2)}}$

$\frac{2}{\mathrm{(x\; -\; 3)}}$ - $\frac{3}{\mathrm{(x\; -\; 2)}}$ = $\frac{\mathrm{5\; -\; x}}{\mathrm{(x\; -\; 3)((x\; -\; 2)}}$

Compare $\frac{\mathrm{5\; -\; x}}{\mathrm{(x\; -\; 3)((x\; -\; 2)}}$ to $\frac{\mathrm{p}}{\mathrm{(x\; -\; 3)((x\; -\; 2)}}$

The numerator p = 5 - x

$\frac{\mathrm{2x}}{1-{\mathrm{x}}^2}$ + $\frac{1}{\mathrm{1\; +\; x}}$ = $\frac{\mathrm{2x}}{1-{\mathrm{x}}^2}$ + $\frac{1}{\mathrm{1\; +\; x}}$

Notes:

1. a² - b² = (a + b)(a - b)

2. 1 - x² = 1² - x² = (1 + x)(1 - x)

$\frac{\mathrm{2x}}{1-{\mathrm{x}}^2}$ + $\frac{1}{\mathrm{1\; +\; x}}$ = $\frac{\mathrm{2x}}{\mathrm{(1\; +\; x)(1\; -\; x)}}$ + $\frac{1}{\mathrm{1\; +\; x}}$

$\frac{\mathrm{2x}}{1-{\mathrm{x}}^2}$ + $\frac{1}{\mathrm{1\; +\; x}}$ = $\frac{\mathrm{1\; x\; 2x\; +\; 1\; x\; (1\; -\; x)}}{\mathrm{(1\; +\; x)(1\; -\; x)}}$

$\frac{\mathrm{2x}}{1-{\mathrm{x}}^2}$ + $\frac{1}{\mathrm{1\; +\; x}}$ = $\frac{\mathrm{2x\; +\; 1\; -\; x}}{\mathrm{(1\; +\; x)(1\; -\; x)}}$

$\frac{\mathrm{2x}}{1-{\mathrm{x}}^2}$ + $\frac{1}{\mathrm{1\; +\; x}}$ = $\frac{\mathrm{2x\; -\; x\; +\; 1}}{\mathrm{(1\; +\; x)(1\; -\; x)}}$

$\frac{\mathrm{2x}}{1-{\mathrm{x}}^2}$ + $\frac{1}{\mathrm{1\; +\; x}}$ = $\frac{\mathrm{x\; +\; 1}}{\mathrm{(1\; +\; x)(1\; -\; x)}}$

Note: x + 1 = 1 + x (the order of addition doesn't matter)

$\frac{\mathrm{2x}}{1-{\mathrm{x}}^2}$ + $\frac{1}{\mathrm{1\; +\; x}}$ = $\frac{\mathrm{1\; +\; x}}{\mathrm{(1\; +\; x)(1\; -\; x)}}$

(1 + x) cancels each other 1 time.

$\frac{\mathrm{2x}}{1-{\mathrm{x}}^2}$ + $\frac{1}{\mathrm{1\; +\; x}}$ = $\frac{1}{\mathrm{1\; -\; x}}$