Question

Solve it fast let me mark u as brainliest

Answer 1

$\frac{1}{(x - 1)(x - 2)} + \frac{1}{(x - 2)(x - 3)} + \frac{1}{(x - 3)(x - 4)} = \frac{1}{6} \\ \frac{(x - 3)(x - 4) + (x - 1)(x - 4) + (x - 1)(x - 2)}{(x - 1)(x - 2)(x - 3)(x - 4)} = \frac{1}{6} \\ \frac{( {x}^{2} - 3x - 4x + 12) + ( {x}^{2} - 4x - x + 4) + ( {x}^{2} - x - 2x + 2) }{(x - 1)(x - 2)(x - 3)(x - 4)} = \frac{1}{6} \\ \frac{3 {x}^{2} - 15x + 18 }{(x - 1)(x - 2)(x - 3)(x - 4)} = \frac{1}{6} \\ \frac{3( {x}^{2} - 5x + 6) }{(x - 1)(x - 2)(x - 3)(x - 4)} = \frac{1}{6} \\ \frac{3(x - 2)(x - 3)}{(x - 1)(x - 2)(x - 3)(x - 4)} = \frac{1}{6} \\ \frac{3}{(x - 1)(x - 4)} = \frac{1}{6} \\ 3 \times 6 = (x - 1)(x - 4) \\ 18 = {x}^{2} - x - 4x + 4 \\ 18 - 4 = {x}^{2} - 5x \\ {x}^{2} - 5x - 14 = 0 \\ {x}^{2} + 2x - 7x - 14 = 0 \\ x(x + 2) - 7(x + 2) = 0 \\ (x - 7)(x + 2) = 0 \\ then \: x - 7 = 0 \\ x = 7 \\ \\ x + 2 = 0 \\ x = - 2 \\ so \: \: x = 7\: and \: - 2$
2)

if a root of the equation x²+px+12 = 0 while the roots of the equation x²+px+q = 0 are same then the value of q is

solution: for this type of questions use options

take option b)4/49
q = 4/49

then x² + px + 4/49 = 0 which has same roots so b²-4ac= 0 used to find p
b² - 4ac = 0
p² - 4*4/49 = 0
p² = 16/49
p = √(16/49)
p = (+ or -)4/7

substitute p in x²+px+12 = 0
x² (+ or -)4/7x +12 = 0 which has no roots so option (b) is wrong

take option (c) 49/4
q = 49/4

then x² + px + 49/4 = 0 which has same roots so b²-4ac= 0 used to find p
p² - 4 * 49/4 = 0
p² -49 = 0
p² = 49
p = (+ or -)7
p = +7 or -7
then x²-7x+49/4 = 0
4x² -28x + 49 = 0 which has roots (2x-7)²

first take p = -7 substitute in x²+px+12=0
x² - 7x +12 = 0 which has (x-3) and (x-4) roots so option (c) is correct

Answer 2

your answer is b -2,-7