4
(ax)(ax) = a2 + 2ax + x2
You notice the first two items, 3bx + 3x factorizes into 3x(b+1). And the last two, by +y factorize into y(b+1) . So (b+1) is a common factor, so you can factorize these two into (b+1)(3x+y)
v=vo + at OR v2=vo2+2ax and one more basic one that I'm forgetting...
The centre is (a, a) and the radius is a*sqrt(2).
x = (-b (+/-) root( b2 - 4ac)) / 2a Sorry about the messy answer, there are no square root symbols or plus-minus symbols. Here is the proof - ax2 + bx + c = 0 --------> multiply by 4a 4a2x2 + 4abx + 4ac = 0 4a2x2 + 4abx = -4ac ----------> add b2 to both sides 4a2x2 + 4abx + b2 = b2 - 4ac -----------> factorise LHS (2ax + b)2 = b2 - 4ac 2ax + b = (+/-) root( b2 - 4ac) 2ax = -b (+/-) root( b2 - 4ac) x = (-b (+/-) root( b2 - 4ac)) / 2a Maths works people.
4
ax2+bx+c=0 Multiply the whole equation by 4a: 4a2x2+4abx+4ac=0 Move the 4ac to the other side: 4a2b2+4abx=-4ac Add b2 to both sides: 4a2b2+4abx+b2=b2-4ac The left side of the equation is like (a+b)2=a2+2ab+b2, with a being 2ab and b being b: (2ax+b)2=b2-4ac Do a square root on both sides of the equation: 2ax+b=√(b2-4ac) Move the b to the other side of the equation: 2ax=-b±√(b2-4ac) Leave only x on the left side of the equation by dividing the right side by 2a: x=(-b±√(b2-4ac))/2a The previous explanation: aX2+bx+c is the same as x=-b (plus or minus the square root) of b2 - 4ac divided by two times a. x=(-b±√(b^2-4ac))/2a x can equal (-b+√(b^2-4ac))/2a ∆ (not delta) (-b-√(b^2-4ac))/2a
42ax = 2ax × 3ax × 7ax
(x - a)(x - a) = x2 -2ax + a2
(ax)(ax) = a2 + 2ax + x2
You notice the first two items, 3bx + 3x factorizes into 3x(b+1). And the last two, by +y factorize into y(b+1) . So (b+1) is a common factor, so you can factorize these two into (b+1)(3x+y)
x = 1 divided by (a - 2.5)
f (x) = ax² + bx + c f ` (x) = 2ax + b = 0 for turning point 2ax = - b x = - b / 2a as required
30-16-27
v=vo + at OR v2=vo2+2ax and one more basic one that I'm forgetting...
The centre is (a, a) and the radius is a*sqrt(2).