Simplification

BODMAS, surds, indices

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Simplification is the process of reducing complex mathematical expressions to their simplest form. In competitive exams, simplification problems test your command over basic arithmetic operations, understanding of operator precedence (BODMAS), and ability to work with fractions, decimals, surds, and indices. These questions form the foundation of quantitative aptitude and require both speed and accuracy.

Key Formulas

BODMAS Rule: Brackets -> Orders (powers/roots) -> Division -> Multiplication -> Addition -> Subtraction\text{BODMAS Rule: Brackets -> Orders (powers/roots) -> Division -> Multiplication -> Addition -> Subtraction}
(a + b)^2 = a^2 + 2ab + b^2\text{(a + b)\textasciicircum{}2 = a\textasciicircum{}2 + 2ab + b\textasciicircum{}2}
(a - b)^2 = a^2 - 2ab + b^2\text{(a - b)\textasciicircum{}2 = a\textasciicircum{}2 - 2ab + b\textasciicircum{}2}
(a + b)(a - b) = a^2 - b^2\text{(a + b)(a - b) = a\textasciicircum{}2 - b\textasciicircum{}2}
(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3\text{(a + b)\textasciicircum{}3 = a\textasciicircum{}3 + 3a\textasciicircum{}2b + 3ab\textasciicircum{}2 + b\textasciicircum{}3}
a^3 + b^3 = (a + b)(a^2 - ab + b^2)\text{a\textasciicircum{}3 + b\textasciicircum{}3 = (a + b)(a\textasciicircum{}2 - ab + b\textasciicircum{}2)}
a^3 - b^3 = (a - b)(a^2 + ab + b^2)\text{a\textasciicircum{}3 - b\textasciicircum{}3 = (a - b)(a\textasciicircum{}2 + ab + b\textasciicircum{}2)}
a^m x a^n = a^(m+n)\text{a\textasciicircum{}m x a\textasciicircum{}n = a\textasciicircum{}(m+n)}
a^m / a^n = a^(m-n)\text{a\textasciicircum{}m / a\textasciicircum{}n = a\textasciicircum{}(m-n)}
(a^m)^n = a^(mn)\text{(a\textasciicircum{}m)\textasciicircum{}n = a\textasciicircum{}(mn)}
a^0 = 1 (a != 0)\text{a\textasciicircum{}0 = 1 (a != 0)}
a^(-n) = 1/a^n\text{a\textasciicircum{}(-n) = 1/a\textasciicircum{}n}
sqrta x sqrta = a\text{sqrta x sqrta = a}
sqrt(a x b) = sqrta x sqrtb\text{sqrt(a x b) = sqrta x sqrtb}
sqrt(a/b) = sqrta/sqrtb\text{sqrt(a/b) = sqrta/sqrtb}
Rationalization: 1/(sqrta + sqrtb) = (sqrta - sqrtb)/(a - b)\text{Rationalization: 1/(sqrta + sqrtb) = (sqrta - sqrtb)/(a - b)}

Key Concepts

BODMAS - Order of Operations

BODMAS dictates the sequence for simplifying expressions: B (Brackets - innermost first), O (Orders - powers and roots), D/M (Division and Multiplication - left to right), A/S (Addition and Subtraction - left to right). Common brackets types: () parentheses, {} braces, [] square brackets. Remember that division and multiplication have equal precedence, as do addition and subtraction. Always work left to right when precedence is equal.

Fractions and Decimals

When simplifying mixed expressions with fractions and decimals, convert all terms to a common format. For addition/subtraction of fractions, find the LCM of denominators. For multiplication, multiply numerators and denominators separately. For division, multiply by the reciprocal. Remember that a/b + c/d = (ad + bc)/(bd) and a/b x c/d = (ac)/(bd). Converting recurring decimals to fractions: 0. a b c = abc/999.

Surds (Irrational Roots)

Surds are irrational numbers expressed as roots (sqrt2, sqrt3, etc.). A surd is in simplest form when the number under the root has no square factors. To simplify sqrtn, factor n into a perfect square and another factor: sqrt12 = sqrt(4x3) = 2sqrt3. Like surds can be combined: 3sqrt2 + 5sqrt2 = 8sqrt2. Unlike surds (sqrt2 + sqrt3) cannot be combined directly. Rationalizing denominators involves eliminating surds from denominators by multiplying numerator and denominator by the conjugate.

Laws of Indices

Indices (exponents) follow specific laws: Product law (a^m x a^n = a^(m+n)), Quotient law (a^m / a^n = a^(m-n)), Power law ((a^m)^n = a^(mn)), Zero law (a^0 = 1), Negative law (a^(-n) = 1/a^n). These laws apply when bases are the same. For different bases, look for common factors or rewrite terms. When dealing with complex expressions, apply the laws systematically from innermost to outermost.

Algebraic Identities in Simplification

Algebraic identities help simplify complex expressions efficiently. Key identities: (a+b)^2 = a^2+2ab+b^2, (a-b)^2 = a^2-2ab+b^2, a^2-b^2 = (a+b)(a-b), a^3+b^3 = (a+b)(a^2-ab+b^2), a^3-b^3 = (a-b)(a^2+ab+b^2). These are particularly useful when simplifying expressions like 47^2 + 2x47x53 + 53^2, which directly fits (a+b)^2 pattern giving (47+53)^2 = 100^2 = 10000.

Modulus and Absolute Value

The modulus |x| represents the absolute value of x, always returning a non-negative result. |x| = x if x >= 0, and |x| = -x if x < 0. When simplifying expressions with modulus: |a - b| = |b - a|, |a x b| = |a| x |b|, |a/b| = |a|/|b|. Nested moduli are solved from innermost to outermost. Remember that |x| = k implies x = k or x = -k.

Solved Examples

Problem 1:

Simplify: 48 / 12 x ( 3 + [5 - {18 / 6 - 2} ] )

Solution:

Step 1: Start with innermost brackets - curly braces { }
Step 2: Inside { }: 18 / 6 - 2 = 3 - 2 = 1
Step 3: Expression becomes: 48 / 12 x ( 3 + [5 - 1] )
Step 4: Inside square brackets [ ]: 5 - 1 = 4
Step 5: Expression becomes: 48 / 12 x ( 3 + 4 )
Step 6: Inside parentheses ( ): 3 + 4 = 7
Step 7: Now: 48 / 12 x 7
Step 8: Division and multiplication left to right: 48 / 12 = 4
Step 9: 4 x 7 = 28
Answer: 28

Problem 2:

Simplify: (0.8 - 0.3 ) / (0.8 x 0.3 )

Solution:

Step 1: Convert recurring decimals to fractions
Step 2: 0.8 = 8/9 (single digit recurring -> digit/9)
Step 3: 0.3 = 3/9 = 1/3
Step 4: Numerator: 8/9 - 1/3 = 8/9 - 3/9 = 5/9
Step 5: Denominator: 8/9 x 1/3 = 8/27
Step 6: Division: (5/9) / (8/27) = (5/9) x (27/8)
Step 7: = (5 x 27) / (9 x 8) = 135 / 72
Step 8: Simplify: 135/72 = 15/8 = 1.875
Answer: 15/8 or 1.875

Problem 3:

If a = 7 + 4sqrt3, find the value of sqrta + 1/sqrta

Solution:

Step 1: First find 1/a = 1/(7 + 4sqrt3)
Step 2: Rationalize: multiply by (7 - 4sqrt3)/(7 - 4sqrt3)
Step 3: 1/a = (7 - 4sqrt3) / (49 - 48) = 7 - 4sqrt3
Step 4: Notice that a + 1/a = (7 + 4sqrt3) + (7 - 4sqrt3) = 14
Step 5: We need sqrta + 1/sqrta, let this be x
Step 6: x^2 = (sqrta + 1/sqrta)^2 = a + 1/a + 2 = 14 + 2 = 16
Step 7: Therefore x = sqrt16 = 4
Answer: 4

Problem 4:

Simplify: [(2^3)^2 x 3 ] / [2 x 3^2]

Solution:

Step 1: Simplify numerator using power law: (2^3)^2 = 2
Step 2: Numerator: 2 x 3 = 64 x 81
Step 3: Denominator: 2 x 3^2 = 16 x 9 = 144
Step 4: Using quotient law: 2 /2 x 3 /3^2 = 2^2 x 3^2
Step 5: = 4 x 9 = 36
Alternative: 2^(6-4) x 3^(4-2) = 2^2 x 3^2 = 4 x 9 = 36
Answer: 36

Tips & Tricks

  • Always follow BODMAS strictly - incorrect bracket order is the most common mistake in simplification.
  • For recurring decimals, remember: 0. a = a/9, 0. a b = ab/99, 0. a b c = abc/999, and 0.a b c = (abc - a)/990.
  • When dealing with complex fractions, simplify step by step rather than trying to solve everything at once.
  • For surd rationalization, remember the pattern: multiply by the conjugate to eliminate roots from denominators.
  • In expressions with indices, look for common bases first before applying any laws - this often reveals shortcuts.
  • Use algebraic identities for quick calculations: (a+b)^2, (a-b)^2, and a^2-b^2 patterns appear frequently in exam questions.
  • When dealing with mixed fractions and decimals, convert to a single format (usually fractions) for easier manipulation.
  • For modulus problems, carefully track the sign of expressions inside - |x| equals x when x>=0 and -x when x<0.

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