Today our topic of discussion is Algebraic Formulae.
Algebraic Formulae
Algebraic Formulae
Any general rule or resolution expressed by algebraic symbols is called Algebraic Formula. In class VII and VIII, algebraic formulae and related corollaries have been discussed. In this chapter, some applications are presented on the basis of that discussion.
Formula 1. (a + b)² = a² + 2ab + b²
Formula 2. (a – b) ²= a ² – 2ab + b²
Remarks: It is seen from Formula 1 and Formula 2 that, adding 2ab or -2ab to a² + b² we get a perfect square, i.e. we get (a + b)² or (a – b)² Substituting -b instead of b in Formula 1 we get Formula 2 / ({a + (- b)}²) = a ² + 2a(- b) + (- b)²
That is, (a – b) ²= a² – 2ab + b²
Corollary 1. a² + b ² = (a + b) ² – 2ab
Corollary 2. a² + b²= (a – b) ² + 2ab
Corollary 3. (a + b) ² = (a – b)² + 4ab
Proof: (a + b) ²= a² + 2ab + b² = a² – 2ab + b² + 4ab = (a – b)² + 4ab
Corollary 4. (a – b)² = (a + b)² – 4ab
Proof: (a – b) ²= a ² – 2ab + b² = a² + 2ab + b² – 4ab = (a + b)² – 4ab
Corollary 5. a² + b ² = ((a + b) ² + (a – b) ²)/2
a²+2ab+b² = (a+b)² a ² – 2ab + b ²= (a – b) ²
adding, 2a ² + 2b ²= (a + b)² + (a – b)² or, 2(a ² + b ²) = (a + b) ² + (a – b) ² (a ² + b ²) = ((a + b)² + (a – b) ²)/2
Corollary 6. ab = ((a + b)/2)²- ((a – b)/2)²
Proof: From Formula 1 and Formula 2, a ²+ 2ab + b² = (a + b) ²
a ² – 2ab + b² = (a – b) ² subtracting, 4ab = (a + b)² – (a – b)² or, ab = ((a + b)²)/4 – ((a – b) ²)/4 hence, ab= ((a + b)/2)² -( dot a -b 2 )²
Remark: By applying the Corollary 6 product of any two quantities can be expressed as the difference of two squares.
Formula 3. a ² – b ² = (a + b)(a – b)
Therefore, the difference of the squares of two expressions = sum of two expressions
x difference of two expressions.
Formula 4. (x + a)(x + b) = x² + (a + b) * x + ab
Therefore, (x + a)(x + b) =x² + (algebraic sum of a and b) x + (the product of a and b)
Extension of formula for square: There are three terms in the expression a + b + c It can be considered the sum of two terms (a + b) and c. Therefore, by applying Formula 1, the square of the expression is,
(a + b + c)² = \{(a + b) + c\}²= (a + b)²+ 2(a + b) * c + c ²
a ²+ 2ab + b² + 2ac + 2bc + c² = a ² + b ² + c ² + 2ab + 2bc + 2c
Formula 5. (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ac
Corollary 7. a² + b² + c² = (a + b + c)² – 2(ab + bc + ac)
Corollary 8. 2(ab + bc + ac) = (a + b + c)² – (a² + b² + c²)
Note: By applying Formula 5, we get,
1) (a + b – c)²= {a + b + (- c)}² = a²+ b² + (- c)² + 2ab + 2b(- c) + 2a(- c) = a ² + b² + c ² + 2ab – 2bc – 2ac
2) (a – b + c)² = {a + (- b) + c} ² = a² + (- b)² + c ² + 2a(- b) + 2(- b) * c + 2ac = a² + b² + c ² – 2ab – 2bc + 2ac
3) (a – b – c)² = {a + (- b) + (- c)}² = a²+ (- b)² + (- c)²+ 2a(- b) + 2(- b)(- c) + 2a(- c) = a ²+ b ² + c²- 2ab + 2bc – 2ac
Example 1.
What is the square of (4x + 5y)
Solution: (4x + 5y)² = (4x)² + 2(4x)(5y) + (5y)² = 16x² + 40xy + 25y²
Example 2. What is the square of (3a – 7b) ?
Solution: (3a – 7b)²= (3a)²- 2(3a)(7b) + (7b)² = 9a² – 42ab + 49b² Example 3. Find the square of 996 by applying the formula of square.
Solution: (996)² = (1000 – 4)² = (1000)² – 2 * 1000 * 4 + 4²= 1000000 – 8000 + 16 = 1000016 – 8000 = 992016
Example 4. What is the square of a + b + c +d7
Solution: (a + b + c + d)²
= {(a + b) + (c + d)}²
= (a + b)² + 2(a + b)(c + d) + (c + d)²
= a ² + 2ab + b² + 2(ac + ad + bc + bd) + c² + 2cd + d²
= a ²+ b²+ c² + d² + 2ab + 2ac + 2ad + 2bc + 2bd + 2cd
Example 5. Simplify: (5x + 7y + 3z)² + 2(7x – 7y – 3z)(5x + 7y + 3z) +(7x-7y – 3z )²
Solution: Let, 5x + 7u + 3z = a and 7x – 7y – 3z = b
∴ Given expression = a² + 2ba + b² = a² + 2ab + b²
= (a + b)²
= {(5x + 7y + 3z) + (7x – 7y – 3z)}² [ substituting the values of a and b
= (5x + 7y + 3z + 7x – 7y – 3z)²
= (12x)² = 144x²
Example 6. If x – y = 2 and xy = 24 what is the value of x + y
Solution: (x + y)²= (x – y)² + 4xy = (2)²+ 4 * 24 = 4 + 96 = 100
therefore x + y = ± √(100) = ± 10
Example 7. If a4+ a ² * b² + b4 = 3 and a² + ab + b² = 3 , what is the value of a² +b² ?
Solution: a4 + a² * b² + b4
= (a²)² + 2a² * b² + (b²)² – a² * b²
= (a² + b²)² – (ab)²
= (a² + b²+ ab)(a² + b² – ab)
= (a²+ ab + b²)(a²- ab + b²)
.. 3 = 3(a² – ab + b²) [substituting the values]
or, a² – ab + b² = 3/3 = 1
Now adding, a² + ab + b² = 3 and a² – ab + b² = 1
we get, 2(a²+ b²) = 4
or, a²+ b² = 4/2 = 2 . a² + b² = 2
Example 8. Prove that, (a + b)4 – (a – b)4 = 8ab(a² + b²)
Solution: (a + b)4 – (a – b)4
= {(a + b)²}² – {(a – b) ²}²
= {(a + b)² + (a – b)²}{(a + b)² – (a – b)²}
= 2(a ² + b² ) * 4ab [ applying Corollary 5 and Corollary 6]
= 8ab(a² + b²)
.. (a + b)4 – (a – b)4 = 8ab(a² + b²)
Example 9. ab + bc +ac?
If a + b + c = 15 and a² + b² + c²= 83 , what is the value of
Solution: First method:
2(ab + bc + ac) = (a + b + c)²- (a²+ b² + c²) = (15)² – 83 = 225 – 83 = 142 therefore ab + bc + ac = 142/2 = 71
Alternative method:
(a + b + c)² = (a² + b² + c²) + 2(ab + bc + ac)
or, (15)²= 83 + 2(ab + bc + ac)
or, 225 – 83 = 2(ab + bc + ac)
or, 2(ab + bc + ac) = 142 therefore ab + bc + ac = 142/2 = 71
Example 10. If a + b + c = 2 and ab + bc + ac = 1 , what is the value of (a + b)² + (b + c)² + (c + a)²?
Solution: (a + b)² + (b + c)² + (c + a)²
= a ² + 2ab + b ² + b² + 2bc + c² + c² + 2ca + a²
= (a² + b²+ c² + 2ab + 2bc + 2ca) + (a² + b² + c²)
= (a + b + c) ²+ (a + b + c) ² – 2(ab + bc + ca)
= (2)² + (2) ² – 2 * 1 = 4 + 4 – 2 = 8 – 2 = 6
Example 11. Express (2x + 3y)(4x – 5y) as the difference of two squares.
Solution: Let, 2x+3y a and 4x – 5y = b
ab = ((a + b)/2)²- ((a – b)/2) ² = ((2x + 3y + 4x – 5y)/2) ² – ((2x + 3y – 4x + 5y)/2)² of a substituting the values
= ((6x – 2y)/2) ²- ((8y – 2x)/2) ²
= {(2(3x – y)/2} ² – {(2(4y – x)/2}² = (3x – y)² – (4y – x) ²
.. (2x + 3y)(4x – 5y) = (3x – y)² – (4y – x) ²
Work:
1) Simplify: (4x + 3y) ² + 2(4x + 3y)(4x – 3y) + (4x – 3y)²
2) If x + y + z = 12 and x² + y ² + z² = 50 find the value of (x – y)² +(y- z)² + (z – x)²
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