Maths

x² - |x| - 12 = 0
Case 1: x ≥ 0, |x| = x
x² - x - 12 = 0
(x-4) (x+3) = 0, x=4 (x=-3 is rejected)
Case 2: x < 0, |x| = -x
x² + x - 12 = 0
(x+4) (x-3) = 0, x=-4 (x=3 is rejected)
Two real solutions: 4 and -4.

x² - 4xy – 5y² = 0
Equation of pair of straight line bisectors is (x²-y²)/ (a-b) = xy/h
(x²-y²)/ (1- (-5) = xy/ (-2)
(x²-y²)/6 = xy/ (-2)
x²-y² = -3xy
x² + 3xy - y² = 0

Kindly go through the solution 

ai+aj+ck,  i+k and ci+cj+bk are co-planar,
|a c; 1 0 1; c b| = 0
a (0-c) - a (b-c) + c (c-0) = 0
-ac - ab + AC + c² = 0
c² = ab
c = √ab

R? → R? -R? , R? → R? -R?
|sinx-cosx, cosx-sinx, 0; 0, sinx-cosx, cosx-sinx; cosx, sinx| = 0
(sinx-cosx)² |1, -1, 0; 0, 1, -1; cosx, sinx| = 0
(sinx-cosx)² (1 (sinx+cosx) + 1 (cosx) = 0
(sinx-cosx)² (sinx + 2cosx) = 0
sin x = cos x
tan x = 1 ⇒ x = π/4
or
sin x = -2cos x
tan x = -2
Not within given range.

∑? [ (-1)? n / 2]
= [8/2] + [-9/2] + [10/2] + [-11/2] + . + [-99/2] + [100/2]
= 4 - 5 + 5 - 6 + . + (-50) + 50
= 4

$CD=\sqrt[]{{\left(10+x^2\right)}^2-{\left(10-x^2\right)}^2}=2\sqrt[]{10}\left|x\right|$

Area 

$=\frac{1}{2}*CD*AB=\frac{1}{2}*2\sqrt[]{10}\left|x\right|\left(20-2x^2\right)$

$\Rightarrow 10-x^2=2x$

3x² = 10

 x = k

3k² = 10