Quadratic Equation Questions
Hello Aspirants, As we all know that Quadratic Equation Questions is a important part of quantitative aptitude section for every competitive exams.Different kind of Quadratic Equation Questions asked in every competitative exam.So here, In this article we will provide Quadratic Equation Quiz.These Quadratic Equation Questions are important for Bank, SSC, SEBI, NABARD, RBI, LIC, and Other state exams. You can attempt these questions & boost your preparation for your examination.
In the Banking exams Quadratic Equation Questions asked in the Prelims as well as Mains exam.There are 5 Quadratic Equation Questions asked in the Prelims exam (Bank).You want to score more in the Quadratic Equation section then you should practice more and more Quadratic Equation.
Quadratic Equation Questions Quiz -4
1.In the following question, two equations are given. You have to solve these equations and find out the values of x and y and determine the relation between them.
I. x2 + 17x + 52 = 0
II. y2 + 16y + 39 = 0
A. x ≥ y
B. x > y
C. x = y or no specific relation can be established
D. x ≤ y
E. x < y
Show Correct Answers
Correct Answer – C. x = y or no specific relation can be established
I. x2 + 17x + 52 = 0
⇒ x2 + 13x + 4x + 52 = 0
Splitting the middle term
⇒ x (x + 13) + 4 (x + 13) = 0
⇒ (x + 4) (x + 13) = 0
So, (x + 4) = 0
⇒ x = – 4
Or, (x + 13) = 0
⇒ x = – 13
II. y2 + 16y + 39 = 0
⇒ y2 + 13y + 3y + 39 = 0
Splitting the middle term
⇒ y (y + 13) + 3 (y + 13) = 0
⇒ (y + 3) (y + 13) = 0
So, (y + 3) = 0
⇒ y = – 3
Or, (y + 13) = 0
⇒ y = – 13
So, when x = – 4, x < y for y = – 3 and x > y for y = – 13
And when x = – 13, x < y for y = – 3 and x = y for y = – 13
∴ so, we can observe that no clear relationship cannot be determined between x and y.
2. In the following question, two equations are given. You have to solve these equations and find out the values of x and y and determine the relation between them.
I. 5x2 + 1 = 6x
II. 16y2 + 1 = 8y
A. x ≥ y
B. x > y
C. x = y or no specific relation can be established
D. x ≤ y
E. x < y
Show Correct Answers
Correct Answer – C. x = y or no specific relation can be established
I. 5x2 + 1 = 6x
⇒ 5x2 – 6x + 1 = 0
⇒ 5x2 – 5x – x + 1 = 0
Now splitting middle term,
⇒ 5x (x – 1) – 1 (x – 1) = 0
⇒ (5x – 1) (x – 1) = 0
So, (5x – 1) = 0
⇒ x = 1/5
Or, (x – 1) = 0
⇒ x = 1
II. 16y2 + 1 = 8y
⇒ 16y2 – 8y + 1 = 0
Now splitting middle term
⇒ 16y2 – 4y – 4y + 1 = 0
⇒ 4y (4y – 1) – 1 (4y – 1) = 0
⇒ (4y – 1) (4y – 1) = 0
Then, y = ¼
So, when x = 1/5, x < y for y = ¼
And when x = 1, x > y for y = ¼
∴ So, we can observe that no clear relationship cannot be determined between x and y
3.In the following question, two equations are given. You have to solve these equations and find out the values of x and y and determine the relation between them.
I. x3 – 120 = 392
II. y3 – 217 = 512
A. x ≥ y
B. y ≥ x
C. x = y or no specific relation can be established
D. x > y
E. y > x
Show Correct Answers
Correct Answer – E. y > x
I. x3 – 120 = 392
⇒ x3 = 392 + 120
⇒ x3 = 512
⇒ x = 8
II. y3 – 217 = 512
⇒ y3 = 512 + 217
⇒ y3 = 729
⇒ y = 9
∴ We can observe that y > x.
4.In the following question, two equations are given. You have to solve these equations and find out the values of x and y and determine the relation between them.
A. x ≥ y
B. y ≥ x
C. x = y or no specific relation can be established
D. x > y
E. x < y
Show Correct Answers
Correct Answer – C. x = y or no specific relation can be established
5. In the following question, two equations are given. You have to solve these equations and find out the values of x and y and determine the relation between them.
I. x2 – 26x = – 5x – 68
II. y2 + 18y = 2y – 48
A. x ≥ y
B. y ≥ x
C. x = y or no specific relation can be established
D. x > y
E. x < y
Show Correct Answers
Correct Answer – D. x > y
I. x2 – 26x = – 5x – 68
⇒ x2 – 26x + 5x + 68 = 0
⇒ x2 – 21x + 68 = 0
Splitting middle term
⇒ x2 – 17x – 4x + 68 = 0
⇒ x (x – 17) – 4 (x – 17) = 0
⇒ (x – 4) (x – 17) = 0
So, x = + 4 or x = + 17
II. y2 + 18y = 2y – 48
⇒ y2 + 18y – 2y + 48 = 0
⇒ y2 + 16y + 48 = 0
⇒ y2 + 12y + 4y + 48 = 0
⇒ y (y + 12) + 4 (y + 12) = 0
⇒ (y + 4) (y + 12) = 0
So, y = – 4 or y = – 12
∴ As, we can observe that both of the values of x are positive and both of the values of y are negative then we can clearly say that x > y
6.In the following question, two equations are given. You have to solve these equations and find out the values of x and y and determine the relation between them.
I. x2 + 96 = 321
II. y + 1296 = 1312
A. x ≥ y
B. y ≥ x
C. x = y or no specific relation can be established
D. x > y
E. x < y
Show Correct Answers
Correct Answer – E. x < y
I. x2 + 96 = 321
⇒ x2 = 321 – 96 = 225
⇒ x = √225
⇒ x = ± 15
II. y +1296 = 1312
⇒ y = 1312 – 1296
⇒ y = 16
So, when x = + 15, x < y for y = 16
And when x = – 15, x < y for y = 16
Hence, we can conclude x < y.
7.In the following question, two equations are given. You have to solve these equations and find out the values of x and y and determine the relation between them.
I. x2 – 16x + 28 = 0
II. y2 – 14y + 13 = 0
A. x ≥ y
B. y ≥ x
C. x = y or no specific relation can be established
D. x > y
E. x < y
Show Correct Answers
Correct Answer – C. x = y or no specific relation can be established
I. x2 – 16x + 28 = 0
⇒ x2 – 14x – 2x + 28 = 0
⇒ x(x – 14) – 2(x – 14) = 0
⇒ (x – 14)(x – 2) = 0
Then, x = + 14 or x = + 2
II.y2 – 14y + 13 = 0
⇒ y2 – 13y – y + 13 = 0
⇒ y(y – 13) – 1(y – 13) = 0
⇒ (y – 13)(y – 1)
Then, y = + 13 or y = + 1
So, when x = + 14, x > y for y = + 13 and x > y for y = + 1
And when x = + 2, x < y for y = + 13 and x > y for y = + 1
∴ So, we can observe that no clear relationship cannot be determined between x and y.
8.In the following question, two equations are given. You have to solve these equations and find out the values of x and y and determine the relation between them.
I. x2 + 6x + 8 = 0
II. y2 + 13y +42 = 0
A. x ≥ y
B. y ≥ x
C. x = y or no specific relation can be established
D. x > y
E. x < y
Show Correct Answers
Correct Answer – D. x > y
I. x2 + 6x + 8 = 0
⇒ x2 + 4x + 2x + 8 = 0
⇒ x(x + 4) + 2(x + 4) = 0
⇒ (x + 4)(x + 2) = 0
Then, x = – 4 or x = – 2
II.y2 + 13y + 42 = 0
⇒ y2 + 6y + 7y + 42 = 0
⇒ y(y + 6) + 7(y + 6) = 0
⇒ (y + 7)(y + 6) = 0
Then, y = – 7 or y = – 6
So, when x = – 4, x > y for y = – 7 and x > y for y = – 6
And when x = – 2, x > y for y = – 7 and x > y for y = – 6
∴ We can clearly observe that x > y.
9.In the following question, two equations are given. You have to solve these equations and find out the values of x and y and determine the relation between them.
I. x2 – 17x + 72 = 0
II. y2 – 23y + 130 = 0
A. x ≥ y
B. y ≥ x
C. x = y or no specific relation can be established
D. x > y
E. x < y
Show Correct Answers
Correct Answer – E. x < y
I. x2 – 17x + 72 = 0
⇒ x2 – 9x – 8x + 72 = 0
⇒ x(x – 9) – 8(x – 9) = 0
⇒ (x – 8)(x – 9) = 0
Then, x = 8 or x = 9
II. y2 – 23y + 130 = 0
⇒ y2 – 13y – 10y + 130 = 0
⇒ y(y – 13) – 10(y – 13) = 0
⇒ (y – 10)(y – 13) = 0
Then, y = 10 or y = 13
As, we can observe that both the values of x are less than both values of y then we can conclude x < y.
10. In the following question, two equations are given. You have to solve these equations and find out the values of x and y and determine the relation between them.
I. 3x2 – 51x + 180 = 0
II. 2y2 – 24y + 64 = 0
A. x ≥ y
B. y ≥ x
C. x = y or no specific relation can be established
D. x > y
E. x < y
Show Correct Answers
Correct Answer – C. x = y or no specific relation can be established
I. 3x2 – 51x + 180 = 0
⇒ x2 – 17x + 60 = 0 [Dividing both sides by 3]
⇒ x2 – 12x – 5x + 60 = 0
⇒ x(x – 12) – 5(x – 12) = 0
⇒ (x – 12)(x – 5) = 0
Then, x = 12 or x = 5
II. 2y2 – 24y + 64 = 0
⇒ y2 – 12y + 32 = 0 [Dividing both sides by 2]
⇒ y2 – 8y – 4y + 32 = 0
⇒ y(y – 8) – 4(y – 8) = 0
⇒ (y – 8)(y – 4) = 0
Then, y = 8 or y = 4
So, when x = 12, x > y for y = 8 and x > y for y = 4
And when x = 5, x < y for y = 8 and x > y for y = 4
∴ So, we can observe that no clear relationship cannot be determined between x and y.
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