→ If ax2 + bx + c = 0, then x = fraction numerator – straight b plus-or-minus square root of straight b squared minus 4 ac end root over denominator 2 straight a end fraction

→ If open parentheses straight x plus 1 over straight x close parentheses = 2, then x = 1

→ If open parentheses straight x plus 1 over straight x close parentheses = –2, then x = –1

open parentheses straight x cubed plus 1 over straight x cubed close parentheses equals open parentheses straight x plus 1 over straight x close parentheses cubed – 3 open parentheses straight x plus 1 over straight x close parentheses

open parentheses straight x cubed minus 1 over straight x cubed close parentheses equals open parentheses straight x minus 1 over straight x close parentheses cubed plus 3 open parentheses straight x minus 1 over straight x close parentheses

→ If open parentheses straight x plus 1 over straight x close parentheses = 1 , then x3 = –1

→ If open parentheses straight x plus 1 over straight x close parentheses = square root of 3, then x6 + 1 = 0

→ If ax + by = m and bx – ay = n, then

      (a2 + b2) (x2 + y2) = (m2 + n2)

→ If (x – a)2 + (y – b)2 + (z – c)2 = 0

      then x – a = 0 ⇒ x = a

              y – a = 0 ⇒ y = b

              z – c = 0 ⇒ z = c

→ If Y = square root of straight x plus square root of straight x plus square root of straight x plus........ space infinity end root end root end root

     and x = n × (n + 1), then y = (n + 1)

→ If Y = square root of straight x – square root of straight x minus square root of straight x minus........ space infinity end root end root end root

     and x = n × (n + 1),  then Y = n

→ If x = fraction numerator 4 square root of ab over denominator square root of straight a plus square root of straight b end fraction, then open parentheses fraction numerator straight x plus 2 square root of straight a over denominator straight x – 2 square root of straight a end fraction plus fraction numerator straight x plus 2 square root of straight b over denominator straight x minus 2 square root of straight b end fraction close parentheses = 2

→ If x = fraction numerator 2 square root of ab over denominator square root of straight a plus square root of straight b end fraction, then open parentheses fraction numerator straight x plus square root of straight a over denominator straight x – square root of straight a end fraction plus fraction numerator straight x plus square root of straight b over denominator straight x minus square root of straight b end fraction close parentheses = 2

→ If ax2 + bx + c = 0 is a quadratic equation,

     and α, β are the roots of this equation

     then

      i)   α + β =  –b/a

      ii) αβ = c/a

      iii) α2 + β2 = (α + β)2 -  2αβ = straight b squared over straight a squared minus fraction numerator 2 straight c over denominator straight a end fraction equals open parentheses fraction numerator straight b squared minus 2 ac over denominator straight a squared end fraction close parentheses

→ If straight a over straight b equals space straight c over straight d

     then fraction numerator straight a plus straight b over denominator straight a – straight b end fraction equals fraction numerator straight c plus straight d over denominator straight c minus straight d end fraction

Concept-1

Same Basic Formulae -

1. (A + B)2 = A2 + B2+ 2AB

2. (A – B)2 = A2 + B2 – 2AB

3. (A + B + C)2 = A2 + B2 + C2 + 2 (AB + BC + CA)

4. (A + B)3 = A3 + B3 + 3AB (A + B)

5. (A – B)3 = A3 – B3 – 3AB (A – B)

6. A3 + B3 = (A + B) (A2 + B2 – AB)

7. A3 – B3 = (A – B) (A2 + B2 + AB)

8. (A + B)2 = (A – B)2 +4AB

9. (A – B)2 = (A + B)2 – 4AB.

10. (A + B + C)3 = A3 + B3 + C3 + 3(A + B) (B + C) (C + A) and if A + B + C = 0  then A3 + B3 + C3 = 3ABC.

11. A3 + B3 + C3 – 3ABC = [(A + B + C) (A2 + B2 + C2 – AB – BC – CA)]

                                            = 1 half left parenthesis straight A space plus space straight B space plus straight C right parenthesis space left square bracket left parenthesis straight A space minus space straight B right parenthesis squared space plus left parenthesis straight B minus straight C right parenthesis squared space plus space left parenthesis straight C minus straight A right parenthesis squared space right square bracket

12. A2 (B +C) + B2 (C + A) + C2 (B + A) + 3ABC = (A + B + C) (AB + BC + CA)

• Law of Indices

1. xa × xb = xa+b

2. straight x to the power of straight a over straight x to the power of straight b      = xa–b

3. (xa)b  = (x)ab

4. open square brackets left parenthesis straight x right parenthesis to the power of straight a close square brackets to the power of space to the power of straight b end exponent equals left parenthesis straight x right parenthesis to the power of straight a to the power of straight b end exponent

5. (xy)a = xa ya

• Rules of i [iota]

(i) The value of i = square root of negative 1 end root

(ii) i2 = – 1

(iii) 1 over straight i equals negative straight i

(iv) It is an imaginary value.

• Laws of divisibility

1. Term (An + Bn) will divisible by (A + B), if n=odd.

2. Term (An + Bn) will never be divisible by (A + B), if n=even.

3. (An – Bn) will always be divisible by (A – B) whether n = odd or even.

4. (An – Bn)  will be divisible by (A + B) if n = even.

Simplification

Order of signs to follow

* BODMAS rule

B → Bracket

O → of

D → Division

M → Multiplication

A → Addition

S → Subtraction

Order of Brackets

* (  ) ← {  } ← [  ]

Vernaculum → When an expression contains Vernaculum (bar),it needed to be solved before all other symbolic expressions.

System of linear equations → 

Let two linear equations are

a1x + b1 y + c1 = 0 and

a2 x + b2 y + c2 = 0

        Conditions                                                    Result

(i) Lines Intersecting at a point                    Unique solution

(ii) Lines parallel to each other                   No solutions

(iii) Coincident lines                                    Infinite solution

Or

(i) If straight a subscript 1 over straight a subscript 2 not equal to space straight b subscript 1 over straight b subscript 2 then there will be only one solution.

(ii) If straight a subscript 1 over straight a subscript 2 equals straight b subscript 1 over straight b subscript 2 equals straight c subscript 1 over straight c subscript 2 then there will be infinite solution.

(iii) If straight a subscript 1 over straight a subscript 2 equals straight b subscript 1 over straight b subscript 2 not equal to straight c subscript 1 over straight c subscript 2 then no solution

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