## IMPORTANT QUESTION WITH TRICK

Published in: Mathematics
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IMPORTANT QUESTION WITH TRICK

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ALGEBRA (SIMPLIFICATION)-I
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Algebra is the part of mathematics in which letters and other general symbols are used to represent numbers and quantities in formulae and equations. Mostly equations are in linear & quadratic form i,e. x+7, x2- 3x+7, x2 -3 , x>3 etc. so basically here, we will discuss all algebraic formulas
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(a + = a2 + b2 + 2ab Example : 78 x 78 + 22 x 22 + 2 x 22 Solution : Let 78 = a & 22 = b 78 x 78 + 22 x 22 +2 x 22 = (78+22)2 10000 Example: If x + y = 12 & xy = 32 then x2 + Y2 Solution : (x + Put x+y = 12 & xy =32 we will get 122 -I-Y2 +2 X 32 or 144 = + Y2 +64 or x2 + Y2 80
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= a2 + b2 - 2ab Example : 63 x 63 + 33 x 33 -2 x 63 x 33 Solution : Let 63 = a & 33 = b 63 = (63 - - 302 900
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Example : 161 x 161 -141 x 141 Solution : Let 161 a & 1 41 161 x 161 -141 x 141 —1612 -1412 Example : Solution : 6040 49 x 49 —13x13 72 49 x 49 —13x13 72 (49 + 72 - 13) b) 302 62x36 = 62/2=31 72
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= a3 + b3 + 3ab(a + b) Example : If a + b — 10 & ab = 21 then a3 + b3 Solution : (a + a 3 + b 3 + 3 ab(a + b) so 103 000 or a3 + b3 - 630 = 370
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= аз - b3 - 3ab(a - Example : The value of 3х0.6хо.4хо.2 Solution : . -0.4х0.4х0.4 - З 0.63 - 0.43 - 3х0.6хо.4(о.6 -0.4) = (0.6+0.4)3 0.23 0.008 The value of . -0,4х0.4х0.4 - з 0.008
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- ab + b2) 13 x 13x 13+ 12x 12 Example : 13x 13 —12x 13 + 12x12 3 Solution : a 2 +122 —ab a2 +b2 —ab Example : Solution : 0.0347 x 0.0347 x 0.0347 +0.96538 0,03472 - 0,0347 x 0,9653 + 0,96532 3 a 2 +b2 —ab - 25 = 0.0347+0.9653 = a 2 +b2 —ab
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(a3 - b3) = (a - b) (a2 + ab + b2) Example : Solution : Example : Solution : 13 x 13 x 13— 12 x 12 x 12 13 x 13 + 12 x 13 + 12 x 12 a 3 —b3 a 2 +b2 + ab —a— a 2 +b2 + ab 2.75 x 2.75 x 2.75 — 2.25 x 2.25 x 2.25 2.75 x 2.75 + 2.75 x 2.25 + 2.25 x 2.25 a3 —b3 -a- a 2 +b2 + ab b - 2.75 - 2.25 - 0.5
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Example: If a + b + c Solution : Put a -k b -k C (9)2 81 _ 9 & a b + + c a = 40 then a 2 + b2 + c2 — 9 & ab -k bC + = 40 + c2 + 40 + c2 + 80
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a3 + + c3 — 3abC = (a + b + c) (a 2 + b2 + c2 — ab - bc - ca) Example: Find a3 -k + c3 — 3abC, if a + b -k c = 9 and Solution : a3 + b3 + e — 3abC = (a + b + c) (a2 + b2 + c2 — ab — bc - ca) If a -k b -k c = 9 and ab -k bC -k — 11, here first of all we have to find out of the value of a2 + b2 + c2 By using this formulae put a + b + c 9 & a b + + c a (9)2 _ a2 + b2 + c2 + Il a2 + b2 + c2 + 22 81 -59 so, a 3 + 133 + c 3 — 3 abc = (9) {59 a b + bc+ c a)} = (9) {59 -l l} = =432 - ab - bc - ca)
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If a + b c = O, then a3 + b3 + c3 = 3abC Example: If (x Then x Solution : let x- 9 = c Here, a 3 + b 3 + c 3 — 3abc so indirectly we can say that so finally, (x- 7 + x- 8 + x- 9) or 3x - 24 = 0 or 3x = 24 orx=8
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Example: If x -y = 2 and x2 + Y2 Solution : — 20, (x + — 2(a2 + b2) Putting the value of x + y = 2 and x2 + Y2 - 20, we Will get (x + + (2)2 2(20) (x + y +4 = 40 so (x + = 40- 4 = 36 (2.697 + 0.498)2 + (2.697 - 0.498)2 Example: (2.6972 + 0.4982) Solution : consider a = 2.697 & b = 0.498 putting in question , we will get 2
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= 4ab 1 Example: 5-+ 3 1 Solution : consider 5 3 136) (0.337 + o. 126)2 Example: 136) 3 16 (0.337 -o. 0.337 xo. 126 Solution : Considera = 0.337 & b = 0.126 Putting in question , we will get ab 4ab 4 ab
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Some more important algebraic formulae: then x = 1 then x , then =N2-2 -2)2 -2 6 -2)3 -3(N2 -2)
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4. If x + — then X x -1 66-1 1 then find the value of x 4 Example: If x + — x 1 , then we know that x6 Solution : x + — x So 48 42 36 -1 36

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