Algebra
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If a + 2√3, find the value of a3 . a6 + 3a3 + 1
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a = 2 + √3
∴ 1 = 1 a 2 + √3 = 2 - √3 = 2 - √3 (2 + √3)(2 - √3) 4 - 3
= 2 - √3
Now,= a³ = 1 a6 + 3a³ + 1 3a³ + 3 + 1/a³
[Dividing numerator and denominator by a³]= 1 a³ + 1 + 3 a³ = 1 
a + 1 
³ - 3 a + 1 + 3 a a = 1 (4)³ - 3(4) + 3 = 1 = 1 (4)³ - 3(4) + 3 55 Correct Option: B
a = 2 + √3
∴ 1 = 1 a 2 + √3 = 2 - √3 = 2 - √3 (2 + √3)(2 - √3) 4 - 3
= 2 - √3
Now,= a³ = 1 a6 + 3a³ + 1 3a³ + 3 + 1/a³
[Dividing numerator and denominator by a³]= 1 a³ + 1 + 3 a³ = 1 a + 1 ³ - 3 a + 1 + 3 a a = 1 (4)³ - 3(4) + 3 = 1 = 1 (4)³ - 3(4) + 3 55
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If a + b + c = 0, then bc + ca + ab = ? bc − a2 ca − b2 ab − c2
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a + b + c = 0
⇒ a = – b – c
⇒ a² = – ab – ac
∴ bc – a² = bc + ab + ac
Similarly,
ca – b² = ca + ab + bc
ab – c² = ab + bc + ca∴ bc + ca + ab bc - a² ca - b² ab - c² = bc + ca + ab ab + bc + ca ab + bc + ca ab + bc + ca = ba + bc + ca = 1 ab + bc + ca Correct Option: A
a + b + c = 0
⇒ a = – b – c
⇒ a² = – ab – ac
∴ bc – a² = bc + ab + ac
Similarly,
ca – b² = ca + ab + bc
ab – c² = ab + bc + ca∴ bc + ca + ab bc - a² ca - b² ab - c² = bc + ca + ab ab + bc + ca ab + bc + ca ab + bc + ca = ba + bc + ca = 1 ab + bc + ca
- Resolve into factors :
(x – 1) (x + 1) (x + 3) (x + 5) + 7
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(x – 1) (x + 5) (x + 1) (x + 3) + 7
= (x2 + 5x – x – 5) (x2 + 3x + x + 3) + 7
= (x2 + 4x – 5) (x2 + 3x + x + 3) + 7
Putting x2 + 4x = y, we have,
Expression = (y – 5) (y + 3) + 7
= y2 – 5y + 3y – 15 + 7
= y2 – 2y – 8
= y2 – 4y + 2y – 8
= y (y – 4) + 2 (y – 4)
= (y + 2) (y – 4)
Now,
y + 2 = x2 + 4x + 2
= x2 + 4x + 2 – 2
= (x + 2)2 – (√2)2
(x + 2 + √2)(x + 2 − √2)
Again, y – 4
= x2 + 4x – 4
= x2 + 4x + 4 – 8
= (x + 2)2 – (2√2)2
= (x + 2 + 2√2)(x + 2 − 2√2)
∴ Factorisation is
= (x + 2 + √2)(x + 2 − √2)
(x + 2 + 2√2)(x + 2 − 2√2)Correct Option: A
(x – 1) (x + 5) (x + 1) (x + 3) + 7
= (x2 + 5x – x – 5) (x2 + 3x + x + 3) + 7
= (x2 + 4x – 5) (x2 + 3x + x + 3) + 7
Putting x2 + 4x = y, we have,
Expression = (y – 5) (y + 3) + 7
= y2 – 5y + 3y – 15 + 7
= y2 – 2y – 8
= y2 – 4y + 2y – 8
= y (y – 4) + 2 (y – 4)
= (y + 2) (y – 4)
Now,
y + 2 = x2 + 4x + 2
= x2 + 4x + 2 – 2
= (x + 2)2 – (√2)2
(x + 2 + √2)(x + 2 − √2)
Again, y – 4
= x2 + 4x – 4
= x2 + 4x + 4 – 8
= (x + 2)2 – (2√2)2
= (x + 2 + 2√2)(x + 2 − 2√2)
∴ Factorisation is
= (x + 2 + √2)(x + 2 − √2)
(x + 2 + 2√2)(x + 2 − 2√2)
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If 2x − 1 = 5, find the value of 27x3 + 1 . 3x 8x3
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2x − 1 = 5 3x
On multiplying both sides by 3/2,3x − 1 = 15 2x 2
On cubing both sides,27x3 − 1 −3.3x. 1 8x3 2x 3x − 1 = 3375 2x 8 ⇒ 27x3 − 1 − 9 × 15 8x3 2 2 = 3375 8 ⇒ 27x3 − 1 = 3375 + 135 8x3 8 4 = 3375 + 270 = 3645 8 8 Correct Option: A
2x − 1 = 5 3x
On multiplying both sides by 3/2,3x − 1 = 15 2x 2
On cubing both sides,27x3 − 1 −3.3x. 1 8x3 2x 3x − 1 = 3375 2x 8 ⇒ 27x3 − 1 − 9 × 15 8x3 2 2 = 3375 8 ⇒ 27x3 − 1 = 3375 + 135 8x3 8 4 = 3375 + 270 = 3645 8 8
- What are the factors of the following expression ?
a2 + 1 − 3a + 13 + 34 : a2 a
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a2 + 1 − 13 a − 1 + 34 a2 a = a − 1 2 + 2 − 3 a − 1 + 34 a a = a − 1 2 − 13 a − 1 + 36 a a Let a − 1 = x a
∴ Expression = x2 – 13x + 36
= x2 – 9x – 4x + 36 = x (x – 9) – 4 (x – 9)
= (x – 4) (x – 9)= a − 1 − 4 a − 1 − 9 a a Correct Option: B
a2 + 1 − 13 a − 1 + 34 a2 a = a − 1 2 + 2 − 3 a − 1 + 34 a a = a − 1 2 − 13 a − 1 + 36 a a Let a − 1 = x a
∴ Expression = x2 – 13x + 36
= x2 – 9x – 4x + 36 = x (x – 9) – 4 (x – 9)
= (x – 4) (x – 9)= a − 1 − 4 a − 1 − 9 a a
