簡單數兩條

1) (x-a)(x-b)=c^2x^2 - (a+b)x + ab - c^2 = 0△ = (a+b)^2 - 4(ab - c^2)= a^2 + 2ab + b^2 - 4ab + 4c^2= a^2 - 2ab + b^2 + 4c^2= (a - b)^2 + (2c)^2since (a - b)^2 >= 0 and (2c)^2 >= 0 ,so △ >= 0 therefore the equation (x-a)(x-b)=c^2 always has real roots.When △ = 0 , i.e.(a - b)^2 = 0 and (2c)^2 = 0
==>When a = b and c = 0the equation should have equal roots.

2) | |x+3| - |x-3| | > 3 |x+3| - |x-3| > 3 or |x+3| - |x-3| < - 3 For x < - 3 :(- x - 3) - (3 - x) >3 or (- x - 3) - (3 - x) < - 3- 6 > 3 or - 6 < 3(rejected) For - 3 =< x < 3 :x+3 - (3-x) > 3 or x+3 - (3-x) < - 3x > 3/2 or x < - 3/2 (Answer)
For x >= 3 :(x+3) - (x-3) > 3 or (x+3) - (x-3) < - 36 > 3 or 6 < - 3 (rejected)

Understand graphically, not a proof.

y=(x-a)(x-b)
y=c^2
always have intersection,
they intersect at one point iff a=b & c=0.