Test: SSC CGL Previous Year Questions: Algebra (2023-18) - 3 - SSC CGL MCQ

(2x - (3/x)) = 2
Squaring both side
4x² + 9/(x²) - 2 × 2 × 3 = 4
4x² + 9/(x²) = 16 (squaring both side)
(16x⁴ + 81/(x⁴)) + 2 × 4 × 9 = 256
(16x⁴ + 81/(x⁴)) = 256 - 72 = 184

a - (12/a) = 1
a² -12 = a ⇒ a² - a - 12 = 0
a² - 4a + 3a -12 = 0
a(a - 4) + 3(a - 4) = 0
(a + 3)(a - 4) = 0 ⇒ a = - 3 and 4
Put a = 4
a² + 16/a² = 16 + 16/16 = 16 + 1 = 17

4x⁴ - 37x² + 9 = 0 dividing both side by x²
Then 4x² + 9/x² = 37 subtracting 12 and taking root from both side
2x - 3/x = 5 taking cube on both side
8x³ - 27/(x³) = 5³ + 3 × 5 × 2 × 3 = 125 + 90 = 215

a² + c² + 17 = 2(a - 8b - 2b²)
a² + c² + 17 - 2a + 16b + 4b² = 0
(a - 1)² + (2b + 4)² + c² = 0
a = 1, b = - 2, c = 0
a³ + b³ + c³ = (1)³+ (-2)³ + 0 = - 7

Short-Trick:
(a + b)² - (a - b)² = 4ab

a² + b² + c² + 216 = 12(a + b - 2c)
(a − 6)² + (b - 6)² + (c + 12)² = 0
a = 6, b = 6, c = - 12

= √180 = 6√5

2x² - 7x + 5 = 0

Cubing both side,

Squaring both side,

Short-Trick: 

x = 4, satisfy

As we know that when a + b + c = 0 the a³ + b³ + c³ = 3abc
So, 
5/2(a³ + b³)(b³ + c³)(c³ + a³)

x⁸ - 34x⁴ + 1 = 0 dividing both side by x⁴
x⁴ + 1/x⁴ = 34, adding 2 and taking root on both side
x² + 1/x² = 6 subtracting 2 and taking root on both side
x - 1/x = 2 taking cube on both side
x³ - 1/(x³) = 2³ + 2 × 3 = 8 + 6 = 14

Taking the square of (3x + y)²
(3x + y)² = 9x² + y² + 6xy
(3x + y)² = 37 + 6 × 2
(3x + y)² = 49 ⇒ 3x + y = 7
Taking cube on both side
27x³ + y³ + 9xy(3x + y) = 343
27x³ + y³ = 343 - 9 x 2 x 7
27x³ + y³ = 217

÷ 

= 0.75 + 0.25 = 1

 on rationalizing

= 12√3 − 20 + 9 − 5√3 = 7√3 − 11
a = 7 and b = 11 ⇒ a + b =18

In the equation:
(a + b - c + d)² - (a - b + c - d)²
Use formula : x² - y² = (x - y)(x + y)
(a + b - c + d)² - (a - b + c - d)² = (a + b - c + d + a - b + c - d)(a + b - c + d - a + b - c + d)
= (2a)[2(b - c + d)]
= 4a(b - c + d)

We know,
x³ - y³ = (x - y)(x² + y² + xy)
27a³ - 2√2b³ = (3a)³ - (√2b)³
= (3a - √2b)(9a² + 2b² + 3√2 ab)

5x + 1/3x = 4
Multiply by 3/5 on both sides
We get:

On square both sides

12x² - 21x + 1 = 0
⇒ 12x² + 1 = 21x ⇒ 4x + 1/3x = 7
On multiplying both sides by (3/4x)

squaring equation (i),
⇒ 9x² + (16x²)^-1 + 2 × 3x × 1/4x = 441/16
⇒ 9x² + (16x²)^-1. = 441/16 - 3/2
⇒ 9x² + 1/(16x²) = 417/16

(a + b − 2)² + (b + c − 5)² + (c + a − 5)² = 0
Square of any number is always positive and their sum can be 0 only when they are themselves zero.
Thus, a + b = 2, b + c = 5 and c + a = 5 
a = 1 , b = 1 , c = 4

b + c = ax, a + c = by, a + b= cz
⇒ a + b + c = a + ax, b + a + c = b + by, c + a + b = c + cz

x + y = 13, y + z = 15, z + x = 16
⇒ 2(x + y + z) = 13 + 15 + 16 = 44
x + y + z = 22
x = 7, y = 6, z = 9
15/54 = 5/18

x + 3y + 2 = 0
x + 3y = - 2
⇒ x³ + 27y³ - 18xy = - 8
⇒ x³ + 27y³ - 18xy + 8 = 0

Given, 

Now,


And

Comparing this with the equation (1) 
⇒ a = 7 and b = 5

Put a = 2 , b = 1 and c = 0
ab(a - b) + bc(b - c) + ca(c - a) = 2 × 1(2 - 1) + 1 × 0 (1 - 0) + 0 × 2 (0 - 2) = 2
Now, Go throw options Choose option (d)
⇒ (b - a)(b - c)(c - a)
= (1 - 2)(1 - 0)(0 - 2) = 2
Condition satisfied.
Similarly you can put these values in other options but only option (d) will satisfy the condition.

Put a = 1 then , x = 2 and y = 0

We know that
If x + 1/x = k, then 
Here, a + 1/a = 3 ⇒ a² + 1/a² = 3² - 2 = 7
 7² - 2 = 47

Given,
x⁴ - 6x² - 1 = 0, x² - 1/(x²) = 6
Cubing both sides

 ...[(a - b)³ = a³ - b³ - 3ab(a - b)]
⇒ x⁶ - 1/(x⁶) = 216 + 18 = 234

⇒ 234 - 5(6) + 5 = 209

Given, 
x + 1/x = (√7)² - 2  = 5
(If x + 1/x - k, then x² + 1/x² = k² - 2)
x³ + 1/x³ = 5³ - 3(5) = 110
If x + 1/x = k, then x³ + 1/(x³) - k³ - 3k)