**Hello Aspirants**. Welcome to Online Quantitative Aptitude Section in AffairsCloud.com. Here we are creating sample questions in **Quadratic Equations** which is common for all the competitive exams. We have included Some questions that are repeatedly asked in bank exams !!!

Follow the link **To solve Quadratic Equations with the help of Number Line**

**x**^{3}– 4x^{2}– 9x + 36 = 0

**y**^{3}+ y^{2}+ y – 3 = 0

A. X > Y

B. X < Y

C. X ≥ Y

D. X ≤ Y

E. X = Y or relation cannot be established**E. X = Y or relation cannot be established**

Explanation:

x^{3}– 4x^{2}– 9x + 36 = 0

x^{2}(x – 4) – 9(x – 4) = 0

(x^{2}– 9)(x – 4) = 0

(x – 3)(x + 3)(x – 4) = 0

x = 3, -3, 4

y^{3}+ y^{2}+ y – 3 = 0

(y – 1)(y^{2}+ 2y + 3) = 0

y = 1; (y^{2}+ 2y + 3) = 0 => No real Solution**x**^{3}+ 10x^{2}+ 27x + 18 = 0

**y**^{3}+ 9y^{2}+ 23y + 15 = 0

A. X > Y

B. X < Y

C. X ≥ Y

D. X ≤ Y

E. X = Y or relation cannot be established**E. X = Y or relation cannot be established**

Explanation:

x^{3}+ 10x^{2}+ 27x + 18 = 0

A^{3}+ Bx^{2}+ Cx + D = 0

D = Product, B = Sum

D = 18; Factors = 1, 2, 3, 6, 9, 18

Cubic equation has three roots, choose three factors(Sum of three factors = B) => 1, 3, 6

(x + 1)(x + 3)(x + 6) = 0

x = -1, -3, -6

y^{3}+ 9y^{2}+ 23y + 15 = 0

A^{3}+ Bx^{2}+ Cx + D = 0

D = Product, B = Sum

D = 15; Factors = 1, 3, 5, 15

Cubic equation has three roots, choose three factors(Sum of three factors = B) => 1, 3, 5

(y + 1)(y + 3)(y + 5) = 0

y = -1, -3, -5**x**^{3}– 4x^{2}+ 5x – 2 = 0

**y³ + 7y**^{2}+ 14y + 8 = 0

A. X > Y

B. X < Y

C. X ≥ Y

D. X ≤ Y

E. X = Y or relation cannot be established**A. X > Y**

Explanation:

x^{3}– 4x^{2}+ 5x – 2 = 0

D = Product, B = Sum

D = 2; Factors = -1, -2

Cubic equation has three roots, choose three factors(Sum of three factors = B) => -1, -1, -2

(x – 1)(x – 1)(x – 2) = 0

x = 1, 1, 2

y³ + 7y^{2}+ 14y + 8 = 0

D = 8; Factors = 1, 2, 4, 8

Cubic equation has three roots, choose three factors(Sum of three factors = B) => 1, 2, 4

(y + 1)(y + 2)(y + 4) = 0

y = -1, -2, -4**x**^{2}– (16)^{2}= (23)^{2}– 56

**y**^{1/3}– 55 + 376 = (18)^{2}

A. X > Y

B. X < Y

C. X ≥ Y

D. X ≤ Y

E. X = Y or relation cannot be established**D. X ≤ Y**

Explanation:

x^{2}– (16)^{2}= (23)^{2}– 56

x^{2}= 729

x = ± 27

y^{1/3}– 55 + 376 = (18)^{2}

y = 3^{3}= 27**√36x + √64 = 0**

**√81y + (4)**^{2}= 0

A. X > Y

B. X < Y

C. X ≥ Y

D. X ≤ Y

E. X = Y or relation cannot be established**A. X > Y**

Explanation:

√36x + √64 = 0

6x + 8 = 0

x = – 1.33

√81y + (4)^{2}= 0

9y + 16 = 0

y = – 1.77**25/√x + 9/√x = 17√x**

**√y/3 + 5√y/6 = 3/√y**

A. X > Y

B. X < Y

C. X ≥ Y

D. X ≤ Y

E. X = Y or relation cannot be established**B. X < Y**

Explanation:

25/√x + 9/√x = 17√x

34 = 17x

x = 2

21√y/18 = 3/√y

y = 18/7 = 2.57**(625)**^{1/4}x + √1225 = 155

**√196y + 13 = 279**

A. X > Y

B. X < Y

C. X ≥ Y

D. X ≤ Y

E. X = Y or relation cannot be established**A. X > Y**

Explanation:

(625)^{1/4}x + √1225 = 155

5x = 155 – 35

5x = 120 => x = 24

√196y + 13 = 279

14y = 266

y = 19**3x – 2y = 10**

**5x – 6y = 6**

A. X > Y

B. X < Y

C. X ≥ Y

D. X ≤ Y

E. X = Y or relation cannot be established**A. X > Y**

Explanation:

3x – 2y = 10 — (1)

5x – 6y = 6 —(2)

From eqn (1) and (2)

x = 6; y = 4**√x – √6/√x = 0**

**y**^{3}– 6^{(3/2)}= 0

A. X > Y

B. X < Y

C. X ≥ Y

D. X ≤ Y

E. X = Y or relation cannot be established**E. X = Y or relation cannot be established**

Explanation:

√x – √6/√x = 0

x = √6 = 2.44

y^{3}– 6^{(3/2)}= 0

y = 6^{(1/2)}

y = √6 = 2.44**√(x + 6) = √121 – √36**

**y**^{2}+ 112 = 473

A. X > Y

B. X < Y

C. X ≥ Y

D. X ≤ Y

E. X = Y or relation cannot be established**C. X ≥ Y**

Explanation:

√(x + 6) = √121 – √36

√(x + 6) = 11 – 6

√(x + 6) = 5

x + 6 = 25

x = 19

y^{2}+ 112 = 473

y^{2}= 361

y = ± 19

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