Maths

I have my gcse exam in a month. Does anyone know any formulas which they don't teach you in gcse's but can be used to easily solve problems?. E.g.) -b/2a gives the turning point of a quadratic equation rather than completing the square.

Reply 1

m2b

14

(root3 / 4 ) x (a^2 ),, (a being the side) gives you the area of an equilateral triangle rather than using 1/2abSinC

Reply 2

karthik02

OP

5

Original post by m2b

(root3 / 4 ) x (a^2 ),, (a being the side) gives you the area of an equilateral triangle rather than using 1/2abSinC

Thank you

(edited 6 years ago)

Reply 3

savage_queen

14

Original post by m2b

(root3 / 4 ) x (a^2 ),, (a being the side) gives you the area of an equilateral triangle rather than using 1/2abSinC

It's basically the same thing...

Reply 4

m2b

14

Original post by savage_queen

It's basically the same thing...

loool

Reply 5

karthik02

OP

5

Original post by m2b

loool

Just trying to be nice. Figured out.

Reply 6

brainmaster

14

Original post by karthik02

I have my gcse exam in a month. Does anyone know any formulas which they don't teach you in gcse's but can be used to easily solve problems?. E.g.) -b/2a gives the turning point of a quadratic equation rather than completing the square.

herons formula;
area = sqrt {S (S-a)(S-b)(S-c)}, where a,b,c are the length of triangle and S is the sum if the three sides divided by 2 and it wil give you the area. This can be helpful since in the syllabus you will be given a triangle with all sides and with the knowledge you learn in class you will use cosine rule to get a missing angle and then use the formula 1/2 ab sin C. But if you know the herons formula then you won't have to use two formulas and errors r reduced and time is conserved :smile:

Aha I remember something in vectors and coordinate geometry.
position vectors - vector AB = B - A.
Vector CD = D - C. but make sure that its position vectors...they r referred from a fixed point.
Finding the length between any two points on a plane with coordinates (x,y) and (x1,y1) - sqrt {(x - x1)^2 + (y - y1)^2}.
and finally a point C dividing a line lets say AB in the ratio p:q will be given by;
(q*A + p*B)/p+q = C

believe me this last formula for a point dividing a line I really helpful in vectors and saves you alot of time (tons of time).

let me know if I helped :smile:

Reply 7

brainmaster

14

Original post by karthik02

I have my gcse exam in a month. Does anyone know any formulas which they don't teach you in gcse's but can be used to easily solve problems?. E.g.) -b/2a gives the turning point of a quadratic equation rather than completing the square.

also adding to your point;
let's say u = f (x) where f (x) is a quadratic equation, then when you complete the square you will see two things;
the minimum or maxiumum Value of y and the value of x for which the maximum or minimum occurs.
let's complete the square for ax^2 + bx + c = 0
a [(x + b/2a)^2 - (b/2a)^2] + C = 0
a [(x + b/2a)^2 - b^2/4a^2] + C = 0
a (x + b/2a)^2 - b^2/4a + C = 0
now the minimum or maximum value of y will be given by -b^2/4a + C.
the value of x for which the maximum or minimum occurs is given by -(b/2a).

but in the system they don't ask for a direct answer...they will first ask you to complete the square and then the next question will be to identify the maximum or minimum.

I'm sure you know how find the coordinates of turning points but sometimes you will have to identify if its maxiumum or minimum point. It is quite easy to do this with a quadratic equation because it only had either a maximum or minimum and we can know this by the general shape, if a < 0 then it has a maximum point since the shape is a upside down "U" and if a>0 then it has a minimum point because the shape is "U".
What if you had a cubic equation? cubic equations have both maxiumum and minimum and also at Times point of inflextion (where gradient is 0 but its not a minimum or neither maximum).
now to confirm if a point is a maximum or minimum in cubic equation we use d^2y/dx^2 (second derivative). like x^3...it's first derivative will be 3x^2 and second derivative will be 6x....
now you replace the value of x you got and see....if the value if greater than 0 then its a minimum point and if the value if less than 0 then its a maximum point. In short;
if d^2y/dx^2 < 0, then its a maximum point.
if d^2y/dx^2 > 0, then its a minimum point.

You won't need this much but it's always good to know it because I've seen questions coming from A levels in this exam and if you know all these then you will save Time and earn full marks. ..might end up getting a hundred :smile:

Reply 8

brainmaster

14

Original post by karthik02

I have my gcse exam in a month. Does anyone know any formulas which they don't teach you in gcse's but can be used to easily solve problems?. E.g.) -b/2a gives the turning point of a quadratic equation rather than completing the square.

sorry for all this posts..I was trying to remember shortcuts 😅;
the sum of an arthmetic series = n/2 (a + L)where a is the first term and l is the last term....it's the same thing as the normal formula - n/2 (2a + d (n-1))
but what if we write it this way - n/2 (a + a + d (n-1))...now the last term of any arthmetic series is given by a + d (n-1) so you can see how the formula n/2 ( a + l) makes sense.

Reply 9

karthik02

OP

5

Original post by brainmaster

sorry for all this posts..I was trying to remember shortcuts 😅;
the sum of an arthmetic series = n/2 (a + L)where a is the first term and l is the last term....it's the same thing as the normal formula - n/2 (2a + d (n-1))
but what if we write it this way - n/2 (a + a + d (n-1))...now the last term of any arthmetic series is given by a + d (n-1) so you can see how the formula n/2 ( a + l) makes sense.

Thanks a lot mate. Much appreciated.

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