SSC CGL Previous Year Questions: Algebra- 2 - SSC CGL

3x – 8 (2 – x) = –19
3x – 16 + 8x = – 19
11x = – 3
∴ x = (-3)/11

By squaring both sides





⇒ 13x² + 13x = 12x² + 12x + 6
⇒ x² + 3x – 2x – 6 = 0
⇒ x (x + 3) – 2 (x + 3) = 0
⇒ (x – 2) (x + 3) = 0
⇒ x – 2 = 0 ∴ x = 2
⇒ x + 3 = 0 ∴ x = – 3

Here,
x + (1/x) = 17

= 1/14 (4913-3x17)
⇒ (1/14) x 4862 = 2431/7

Here,

then,

According to question,
x - y - √18 = -1 ...(i)
x + y - 3√2 = 1 ...(ii)
From equation (i) and (ii),
We get,
x = 3√2 and y = 1
then,
12xy (x² – y²)
⇒ 12x3√2 x1[(3√2)² - (1)² ]
⇒ 36√2 x 17 = 612√2

Here,
(x – y) = 7
then,
∴ a³ – b³ = (a – b) (a² + ab + b²)
(x – 15)³ – (y – 8)³ = ?
⇒ (x – 15 – y + 8) [(x –15)² + (x – 15) (y – 8) + (y – 8)²]
⇒ (x – y – 7)[ (x – 15)² + (x – 15) (y – 8) + (y – 8)²]
⇒ (7 – 7) [(x – 15)² + (x – 15) (y – 8) + (y – 8)²]
⇒ 0 × [(x – 15)² + (x – 15) (y – 8) + (y – 8)²]
⇒ 0

∵ If x – 2 = 0
∴ x = 2
then,
x² + k₁x + k₂ = 0
(2)² + k₁ × 2 + k₂ = 0
2k₁ + k₂ = – 4 ...(i)
If x + 3 = 0
∴ x = –3
then,
x² + k₁x + k₂ = 0
(–3)² + k₁x – 3 + k₂ = 0
∴ 3k₁ – k₂ = 9 ...(ii)
From equation (i) and (ii),
we get k ₁ = 1 and k₂ = – 6

According to question,
a + b + c = 27
(a – 7) + (b – 9) + (c – 11) = 27 – 7 – 9 – 11
(a – 7) + (b – 9) + (c – 11) = 0
∴ a + b + c = 0
then
a³ + b³ + c³ – 3abc = 0
∴ Required answer = 0.

x = 6 + 2√6
Subtraction by 1 in both side.
x -1 = 6 + 2√6 - 1
x -1 = 5 + 2√6 ⇒ 3 + 2√6


Now,

According to question,

⇒ x² + 1 = 2x
⇒ x² – 2x + 1 = (x – 1)²
∴ x – 1 = 0 ∴ x = 1
∴ x⁶⁴ + x¹²¹ = 1 + 1 = 2.

x + (1/x)= 4
then,
x⁶ + (1/x⁶) = ?
Let x + (1/x) = a
∵
⇒ ((4)³ – 3 × 4 )² – 2
⇒ (64 – 12)² – 2 = (2704 – 2) = 2702

Here,

If a + b + c = 0 then a³ + b³ + c³ = 3abc
let, a = (p – q), b = (q – r), c - (r – p)
than, a + b + c = p – q + q – r + r – p = 0
∴ (p – q)³ + (q – r)² + (r – p)³
= 3(p – q) (q – r) (r – p) = 0

x = 3^1/3 – 3^–1/3
Cubing on both sides


3x³ + 9x = 8

x³ + y³ + z³ – 3xyz = 1/2(x + y + z)
[(x – y)² + (y – z)² + (z – x)²]
= 1/2(222 + 223 + 225)
[(222 – 223)² + (223 – 225)² + (225 – 222)²]
⇒ 1/2[670][1+4+9] = 335 [14] = 4690

Cubing both Sides

⇒ 4x² + 1 = 5x

 (Cubing both sides)
(x + 2)³ = x² + 11
x³ + 8 + 6x² + 12x = x² + 11
x³ + 5x² + 12x = 3

x⁴ + y⁴ – 2 x²y²
= (x² – y²)²

Dividing eq by 2
 Cubing both sides 

(x – 2) (x – p) = x² – ax + b
x² + ( – 2 – p)x + (– P) (–2) = x² – ax + b
– (2 + p) = – a 
2 = a – p

x + y + z = 1, 

xy + yz + zx = – 1
(x + y + z)² = 1
x² + y² + z² + 2xy + 2yz + 2zx = 1
x² + y² +z² + 2(–1) = 1
x² + y² +z² = 3
x³ + y³ + z³ – 3xyz
= (x + y + z) (x² + y² +z² + xy – yz – zx)
x³ + y³ + z³ – 3(–1) = (1) [(3 – (–1)]
x³ + y³ + z³ + 3 = 4
= 4 – 3 = 1

Base side of isosceles Δ = 2x – 2y + 4z
Perimeter = 4x – 2y + 6z
Remaining two sides are = (P-B)/2

x² – 3x + 1 = 0
Dividing Equation by x
x - 3 + (1/x) = 0
x + (1/x) = 3

(a – b) = 2, a³ – b³ = 56
(a – b)³ = a³ – b³ – 3ab (a – b)
8 = 56 – 3ab (2)
– 48 = – 6ab
∴ ab = 8
(a – b)² = a² + b² – 2ab
4 = a² + b² – 16
20 = a² + b²

Now

By adding both equations

 dividing Eq. by 2
 Squaring on both sides

Given x = 2
then x³ + 27x² + 243x + 631
= 8 + 108 + 486 + 631
= 1233

Given p = 99
then p(p² + 3p + 3)
= p((p + 1)² + p + 2)
= 99 ((99 + 1)² + 101)
= 99 × (10000 + 101)
= 999999

5x +9y = 5 ...(i)
125x³ + 729y³ =120 ...(ii)
Now cube both sides of equation (i), we get
⇒ (5x + 9y)³ = (5)³
⇒ 125x³ + 729y³ +135xy(5x + 9y) =125
⇒ 125x³ + 729y³ +135xy x 5 =125
Now, put the value of equation (i)
120 + 135 xy × 5 = 125
⇒ 135 × 5xy = 125 – 120
⇒ 135 × 5xy = 5

Given that

On cubing both sides we get

Given fraction
x⁸ -1 = (x⁴ +1)(x² +1)(x -1)(x +1)
x⁴ + 2x³ - 2x -1 = (x -1)(x +1)³
HCF of Given fractor
= (x -1)(x +1) = x² -1