Here is an intuitive method for working this polynomial problem, which can be solved as easily as adding apples to apples and Oranges to oranges. Think of the same-named variables and variable pairs as different "things." In this example, each polynomial contains none or one or more of these three different things: "ax", "by", and "c". So all we need to do is add (or subtract) up the "ax"s, then the "by"s, and then the "c"s and write it down. ax: 1 in the first polynomial + 2 in the second + none (0) in the third = 3 "ax"s. by: 1 in the first polynomial - 3 in the second +1 in the third = -1 "by"s. c: 1 in the first polynomial + 1 in the second - 1 in the third = 1 "c". The answer is the polynomial made up of the three results above: 3ax - 1by + 1c = 3ax - by + c
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.
(x - a)(x - a) = x2 -2ax + a2
f (x) = ax² + bx + c f ` (x) = 2ax + b = 0 for turning point 2ax = - b x = - b / 2a as required
Find what you can divide both 6ax and 9a by.They can both be divided by 3, leaving 2ax and 3a.These can both be divided by a, leaving 2x and 3.So 6ax + 9a = 3a (2x + 3)The factors of 6ax + 9a are 3a and (2x + a)
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
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.
The centre is (a, a) and the radius is a*sqrt(2).
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Think of it in reverse - trying to get it in the form of (x + a)2 This equals x2 + 2ax + a2 Compare the coefficients&constants to those of the original equation: x2 + 8x + c = x2 + 2ax + a2 2a = 8 and c = a2 a = 4, so c = 16
42ax = 2ax × 3ax × 7ax
(ax)(ax) = a2 + 2ax + x2
(x - a)(x - a) = x2 -2ax + a2
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
2x²-x=0.5 is more manageable in the standard quadratic form : 2x2 - x - 0.5 = 0double this to get rid of that pesky 0.5 : 4x2 - 2x -1 = 0The quadratic formula solves ax2 + bx + c =0 using x = (-b +/- (b2-4ac) 0.5 ) / 2ax equals [minus b plus or minus (square root of {b squared minus 4ac})] all over 2aputting in a=4, b=-2, c=-1 gives (2 +/- (4 + 16)0.5 ) / 8which reduces to( 1 +/- (5)0.5) / 4this evaluates to x= .809 or -.309Notice that plus or minus square root (b2 - 4ac) usually produces two different solutions. Equations in the second degree always have two solutions; if the quadratic is such that b2 equals 4ac the formula seems to give only one solution. Don't worry about this; there are still two solutions, they just happen to be identical!
v=vo + at OR v2=vo2+2ax and one more basic one that I'm forgetting...
Find what you can divide both 6ax and 9a by.They can both be divided by 3, leaving 2ax and 3a.These can both be divided by a, leaving 2x and 3.So 6ax + 9a = 3a (2x + 3)The factors of 6ax + 9a are 3a and (2x + a)