IMPORTANT MATHS FORMULAE FOR CLASS X BOARD EXAM 2018 IGNORE MAAT KARNA DUM KE PADH KE JANA 28 TARIKH KO MATH EXAM HE BEST OF LUCK TO ALL MY FRIENDS.
Algebra:
(a + b)2 = a2 + 2ab + b2
(a – b)2 = a2 – 2ab + b2
(a + b) (a – b) = a2 – b2
(x + a)(x + b) = x2 + (a + b)x + ab
(x + a)(x – b) = x2 + (a – b)x – ab
(x – a)(x + b) = x2 + (b – a)x – ab
(x – a)(x – b) = x2 – (a + b)x + ab
(a + b)3 = a3 + b3 + 3ab(a + b)
(a – b)3 = a3 – b3 – 3ab(a – b)
(x + y + z) 2 = x2 + y2 + z2 + 2xy + 2yz + 2xz
(x + y – z) 2 = x2 + y2 + z2 + 2xy – 2yz – 2xz
(x – y + z)2 = x2 + y2 + z2 – 2xy – 2yz + 2xz
(x – y – z)2 = x2 + y2 + z2 – 2xy + 2yz – 2xz
x3 + y3 + z3 – 3xyz = (x + y + z)(x2 + y2 + z2 – xy – yz -xz)
x2 + y2 = 1212 [(x + y)2 + (x – y)2]
(x + a) (x + b) (x + c) = x3 + (a + b +c)x2 + (ab + bc + ca)x + abc
x3 + y3 = (x + y) (x2 – xy + y2)
x3 – y3 = (x – y) (x2 + xy + y2)
x2 + y2 + z2 -xy – yz – zx = 1212 [(x-y)2 + (y-z)2 + (z-x)2]
Linear Equation:
Linear equation in one variable ax + b = 0, x = – −ba−ba
Quadratic Equation: ax2 + bx + c = 0 x = −b±b2−4ac√2a−b±b2−4ac2a
Discriminant D = b2 – 4ac
Trial and error method
In step 2 : take a product of ‘ Common terms’ as their HCF.
In step3 : Take the product of All the terms, Omit the HCF value which gives you the value of LCM.
Product of LCM* HCF=Product of the two polynomials
Note: If cubical expression is given, it may be factorized by using ‘trial and error’ method.
Remainder theorem:
If (x-2) is a factor of the given expression, then take x-2 = 0 , therefore x = 2 , then substitute this value in p(x) = 5 x*x + 3 x-6
P(2) : 5(2*2) + 3(2)-6 =0 (here taking = 0 is very important. If not taken, answer can’t be found)
If (x-2) leaves a remainder of 4
P(2) : 5(2*2) + 3 (2)-6 =4 ( Here taking = 4 is very important. If not taken, answer can’t be found)
Linear equations in two variables
If pair of linear equation is : a1 + b1y +c1 =0 and a2x + b2 y + c2=0
Then nature of roots/zeroes/solutions :
i. If a1/a2 is not equal to b1/b2 then, system has unique solution, is consistent OR graph is two intersecting lines
ii. If a1/a2 = b1/b2 is not equal to c1/c2 , then system has no solution, is inconsistent OR graph is parallel lines.
iii. If a1/a2=b1/b2=c1/c2, then system has infinite solution, is consistent OR graph are coincident lines.
Quadratic Equations
Note: To find the value of ‘x’ you may adopt either ‘splitting the middle term’ or ‘formula method’.
X=( -b +-(D)^.5)/2a (where D= b*b – 4ac) Hence x= (-b+- (b*b- 4ac)^.5)/2a
Sum of the roots = -b/a & Product of roots= c/a
If roots of an equation are given, then :
Quadratic equation : x*x – (Sum of roots).x + (product of the roots) =0
If Discriminant > 0, then the roots are Real & unequal or unique, lines are intersecting.
Discriminant = 0, then the roots are real & equal , lines are coincident.
Discriminant< 0 , then the roots are imaginary (not real), parallel lines
Triangles:
If two triangles are similar then, ratio of their sides are equal.
i.e if triangle ABC~ triangle PQR then AB/PQ = BC/QR = AC/PR
If triangle ABC ~ triangle PQR then (Area of triangle ABC)/ Area of triangle PQR) = (side*side) /(side*side) = ( AB*AB) /( PQ*PQ) =( BC*BC)/( QR*QR) = (AC*AC)/ (PR*PR)
Trigonometry:
Wherever ‘square’ appears think of using the identities
sin^2(x) + cos^2(x) = 1
tan^2(x) + 1 = sec^2(x)
cot^2(x) + 1 = csc^2(x)
tan(x+ y) = (tan x +tan y) / (1 – tan x tan y)
sin(2x) = 2 sin x cos x
cos(2x) = cos^2(x) – sin^2(x) = 2 cos^2(x) – 1 = 1 – 2 sin^2(x)
tan(2x) = 2 tan(x) and many more are there if u want then I will send u an URL thank-you best of luck?????????
Posted by Rakesh Naik
6 years, 9 months ago
Sadhvi Deshpande 6 years, 9 months ago
1Thank You