In this article we will discuss the detailed explanation of Quadratic Equation with the help of solved examples, MCQs and quizzes of Quadratic Equation by Smartkeeda .
Quadratic Equation Questions for Competitive Exams
Quadratic Equation comes in the section of Quantitative Aptitude which is very important topic for the students who are willing to appear in the banking exams. Everyday as the popularity of Banking Exams is increasing among the students it is also affecting the level of exams which is conducting as a result the level of each topic in these exams becoming difficult day by day but in Quadratic Equation it seems difficult and most time taking question but once you learnt the right strategy, speed and accuracy or follow the instructions by Smartkeeda then Quadratic Equation section will help you to score more marks in your exam. The practice of new pattern quizzes and tricks provided by Testzone of quadratic equation will help you to score well.New Pattern Quadratic Equation PDF:
In New Pattern Quadratic Equation we will learn how to solve any Quadratic Equation which is given in difficult form like coefficient of any variable is given with root value.These kind of New Pattern Quadratic Equation are becoming very common as a result they are continuously being asked in high level exams and they are based on root based quadratic Equations and general Quadratic Equation. In simple words we can say that New pattern Quadratic Equation can be defined as the Quadratic equation in which the coefficient of X or Y in the value of root.
If you are not clear with the definition shown above we will make it clear to you with the help of suitable examples regarding them which will clear your all doubts regarding New Pattern Quadratic Equation.
PDF Set 1 | PDF Set 5 | PDF Set 9 | PDF Set 13 |
PDF Set 2 | PDF Set 6 | PDF Set 10 | PDF Set 14 |
PDF Set 3 | PDF Set 7 | PDF Set 11 | PDF Set 15 |
PDF Set 4 | PDF Set 8 | PDF Set 12 | PDF Set 16 |
Quadratic Equation for Bank Exam:
Quadratic Equation questions with explanations will help the aspirants to score more marks in Banking exams like IBPS Clerk, IBPS PO, SBI PO, SBI Clerk and other competitive exams which also increases the chances of being selected in these exams. Smartkeeda guides and provides you to lots of fully solved quadratic equation practice questions and answers with explanation while on other hand Testzone helps Candidates preparing for banking and other competitive exams can access their ability here , accuracy and speed through quadratic equation quizzes.Quadratic Equation for Bank PO Exam like SBI PO, IBPS PO, LIC AAO:
The SBI PO and LIC AAO are the examinations which are very popular in India among students. every year millions of students enroll in theses exams but do you know only few of them succeed do you know why? Because these exams are not easy. These exams need solid preparation speed strategy and accuracy on all the topics across all the sections. And one of the important topic which is asked in these exams is Quadratic Equation keep one thing in mind not to see this topic as lightly as it seems.Let discuss Quadratic equation in detail with the help of examples :
1. I. x^{2} – 17.5x + 69 = 0
II. 2y^{2} – 25y + 78. = 0
II. 2y^{2} – 25y + 78. = 0
A. if x > y
B. if x ≤ y
C. if x ≥ y
D. if x < y
E. if x = y or relationship between x and y can't be established
Explanation :
I. x^{2} – 17.5x + 69 = 0
x^{2} – 11.5x – 6x + 69 = 0
x(x – 11.5) – 6(x – 11.5) = 0
(x – 11.5)(x – 6) = 0
x = 11.5, 6
II. 2y^{2} – 25y + 78 = 0
2y^{2} – 13y – 12y + 78 = 0
y (2y – 13) – 6(2y – 13) = 0
(2y – 13)(y – 6) = 0
y = 6.5, 6
For x = 11.5 and y = 6.5, 6 x > y
For x = 6, and y = 6.5, 6 x ≤ y
Therefore, relationship can’t be established
Hence, option E is correct.
Hence, option E is correct.
2. I. x^{2} + 6x – 112 = 0
II. y^{2} + 22y + 112 = 0
II. y^{2} + 22y + 112 = 0
A. if x > y
B. if x ≤ y
C. if x ≥ y
D. if x < y
E. if x ≤ y or no relationship can be established between x and y.
Explanation :
I. x^{2} + 6x – 112 = 0
x^{2} + 14x – 8x – 112 = 0
x (x + 14) – 8(x + 14) = 0
(x + 14)(x – 8) = 0
x = 8, – 14
II. y^{2} + 22y + 112 = 0
y^{2} + 8y + 14y + 112 = 0
y(y + 8) + 14(y + 8) = 0
(y + 8)(y + 14) = 0
y = – 8, – 14
For, x = – 14 and y = – 8
x < y
For, x = – 14 and y = – 14
x = y
For, x = – 14 and y = – 14
x = y
But for x = 8 and y = – 8 and – 14
x > y
Therefore, relationship can’t be established
Hence, option E is correct.
3. I. 2x^{2} + 11√3x + 45 = 0
II. y^{2} + 8√3y + 45 = 0
II. y^{2} + 8√3y + 45 = 0
A. if x > y
B. if x ≤ y
C. if x ≥ y
D. if x < y
E. if x = y or relationship between x and y can't be established
Explanation :
I. 2x^{2} + 11√3x + 45 = 0
2x^{2} + 6√3x + 5√3x + 45 = 0
2x(x + 3√3) + 5√3 (x + 3√3) = 0
(2x + 5√3)(x + 3√3) = 0
x = – 3√3, – | 5 | √3 |
2 |
II. y^{2} + 8√3y + 45 = 0
y^{2} + 5√3y + 3√3y + 45 = 0
y(y + 5√3) + 3√3 (y + 5√3) = 0
(y + 5√3)(y + 3√3) = 0
y = – 5√3, – 3√3
For x = – 3√3 and y = – 3√3 x = y
For x = – 3√3 and y = – 5√3 x > y
For x = – | 5 | √3 and y = – 5√3, – 3√3 x > y |
2 |
Therefore, x ≥ y
Hence, option C is correct.
4. I. x^{2} + 9x + 20 = 0
II. y^{2} + y – 12 = 0
II. y^{2} + y – 12 = 0
A. if x > y
B. if x ≤ y
C. if x ≥ y
D. if x < y
E. if x = y or relationship between x and y can't be established
Explanation :
I. x^{2} + 9x + 20 = 0
x^{2} + 5x + 4x + 20 = 0
x(x + 5) + 4(x + 5) = 0
(x + 4)(x + 5) = 0
x = – 4, – 5
II. y^{2} + y – 12 = 0
y^{2} + 4y – 3y – 12 = 0
y (y + 4) – 3(y + 4) = 0
(y + 4)(y – 3) = 0
y = – 4, 3
For x = – 4 and y = – 4, x = y
For x = – 4, y = 3 , x < y
For x = – 5, y = – 4 or 3 , x < y
Therefore, x ≤ y
Hence, option B is correct.
5. I. 3x^{2}– 243 = 0
II. 12y^{5} + 110y^{4} = 0
II. 12y^{5} + 110y^{4} = 0
A. if x > y
B. if x ≤ y
C. if x ≥ y
D. if x < y
E. if x = y or relationship between x and y can't be established
I. 3x^{2} – 243 = 0
3x^{2} = 243, x^{2} = 81 , x = ± 9
II. 12y^{5} + 1100y^{4} = 0
12y^{5} = –110y^{4},
Hence, option E is correct.y = – | 110 | = – 9.16 , 0 |
12 |
For x = 9, and y = – 9.16 , 0 x > y
For x = – 9, and y = – 9.16, x > y
For x = – 9, and y = 0 x < y
For x = – 9, and y = 0 x < y
Therefore, relationship cannot be established
Quadratic Equation for Bank Clerk Exam like SBI Clerk, RBI Assistant, LIC, IBPS Clerk:
Quadratic Equation is an important topic for banking exams like SBI Clerk Pre, SBI Clerk mains, LIC Assistant etc. In these exams Quadratic Equation carries value of 4 to 5 marks from this topic. if you prepared well at Testzone you can easily score these 5 marks with help of some tricks and explanation by Smartkeeda at Testzone. Let us discuss important and various questions which can be asked in Banking Exams.
1. I. 6x^{2} – 37x – 35 = 0
II. 54y^{2} + 87y + 35 = 0
II. 54y^{2} + 87y + 35 = 0
A.x > y
B. x < y
C. x ≥ y
D. x ≤ y
B. x < y
C. x ≥ y
D. x ≤ y
E. if x = y or relationship between x and y can't be established
Explanation :
I. 6x^{2} – 37x – 35 = 0
6x^{2} + 5x – 42x – 35 = 0
x(6x + 5) – 7(6x + 5) = 0
(x – 7)(6x + 5) = 0
II. 54y^{2} + 87y + 35 = 0
54y^{2} + 42y + 45y + 35 = 0
6y(9y + 7) + 5(9y + 7) = 0
(6y + 5)(9y + 7) = 0
Hence, option E is correct.
I. 6x^{2} – 37x – 35 = 0
6x^{2} + 5x – 42x – 35 = 0
x(6x + 5) – 7(6x + 5) = 0
(x – 7)(6x + 5) = 0
x = 7, – | 5 |
6 |
II. 54y^{2} + 87y + 35 = 0
54y^{2} + 42y + 45y + 35 = 0
6y(9y + 7) + 5(9y + 7) = 0
(6y + 5)(9y + 7) = 0
y = – | 5 | , – | 7 |
6 | 9 |
Therefore, for x = 7 or – | 5 | and y = – | 5 | or – | 7 |
6 | 6 | 9 |
Hence, option E is correct.
2. I. 8x^{2} – 18√3x + 27 = 0
II. 15y^{2} – 14√3y + 9 = 0
II. 15y^{2} – 14√3y + 9 = 0
A.x > y
B. x < y
C. x ≥ y
D. x ≤ y
B. x < y
C. x ≥ y
D. x ≤ y
E. if x = y or relationship between x and y can't be established
Explanation :
I. 8x^{2} – 18√3x + 27 = 0
8x^{2} – 12√3x – 6√3x + 27 = 0
4x(2x – 3√3) – 3√3 (2x – 3√3) = 0
(2x – 3√3)(4x – 3√3) = 0
x = | 3 | √3, | 3 | √3 |
4 | 2 |
II. 15y^{2} – 14√3y + 9 = 0
15y^{2} – 9√3y – 5√3y + 9 = 0
3y(5y – 3√3) – √3 (5y – 3√3) = 0
(3y – √3)(5y – 3√3) = 0
y = | 1 | √3, | 3 | √3 |
3 | 5 |
For x = | 3 | √3, | 3 | √3 and y = | 1 | √3, | 3 | √3 |
4 | 2 | 3 | 5 |
x > y
Hence, option A is correct.
3. I. 9x^{2} – 39x + 40 = 0
II. 9y^{2} – 30y + 16 = 0
II. 9y^{2} – 30y + 16 = 0
A.x > y
B. x < y
C. x ≥ y
D. x ≤ y
B. x < y
C. x ≥ y
D. x ≤ y
E. if x = y or relationship between x and y can't be established
Explanation :
9x^{2} – 24x – 15X + 40 = 0
3x(3x – 8) – 5(3x – 8) = 0
(3x – 8)(3x – 5) = 0
x = | 8 | , | 5 |
3 | 3 |
II. 9y^{2} – 30y + 16 = 0
9y^{2} – 6y – 24y + 16 = 0
3y(3y – 2) – 8 (3y – 2) = 0
(3y – 8)(3y – 2) = 0
y = | 8 | , | 2 |
3 | 3 |
For x = | 8 | and y = | 2 |
3 | 3 |
x > y
For x = | 5 | and y = | 8 |
3 | 3 |
x < y
Therefore, relationship can’t be established
Hence, option E is correct.
4. I. 10x^{2} + 13x – 77 = 0
II. 8y^{2} + 45y + 63 = 0
A.x > y
B. x < y
C. x ≥ y
D. x ≤ y
B. x < y
C. x ≥ y
D. x ≤ y
E. if x = y or relationship between x and y can't be established
Explanation :
I. 10x^{2} + 13x – 77 = 0
10x^{2} + 35x – 22x – 77 = 0
5x(2x + 7) – 11(2x + 7) = 0
(5x – 11)(2x + 7) = 0
x = | 11 | , – | 7 |
5 | 2 |
II. 8y^{2} + 45y + 63 = 0
8y^{2} + 24y + 21y + 63 = 0
8y(y + 3) + 21(y + 3) = 0
(8y + 21)(y + 3) = 0
y = – | 21 | , – 3 |
8 |
For x = – | 7 | , and y = – 3 |
2 |
x < y
For x = – | 11 | , and y = -3 |
5 |
x > y
Therefore, relationship can’t be established
Hence, option E is correct.
5. I. x^{2} + 12x + 35 = 0
II. y^{2} + 9y + 20 = 0
II. y^{2} + 9y + 20 = 0
A.x > y
B. x < y
C. x ≥ y
D. x ≤ y
B. x < y
C. x ≥ y
D. x ≤ y
E. if x = y or relationship between x and y can't be established
Explanation :
As a Conclusion, these are the basic concepts of attempting Banking Exams's quadratic equations questions. The main challenge of the Quadratic equations is to check the kind of question presented which allows you to productively approach and solve it. However, that can only come with practice, practice and practice because as we all heard the fact "Practice Makes Man Perfect", and hence, keep practicing Quadratic Equations questions regularly to gives you intelligence among this topic.
In this article we have covered given below queries:
All the Best
Regards
Team Smartkeeda
I. x^{2} + 12x + 35 = 0
x^{2} + 7x + 5x + 35 = 0
x(x + 7) + 5(x + 7) = 0
(x + 5)(x + 7) = 0
x = – 7, – 5
II. y^{2} + 9y + 20 = 0
y^{2} + 4y + 5y + 20 = 0
y(y + 4) + 5(y + 4) = 0
(y + 5)(y + 4) = 0
y = – 5, – 4
For x = – 7 and y = – 5 or – 4
x < y
For x = – 5 and y = – 5, x = y
For x = – 7 and y = – 4
Therefore, x ≤ y
Hence, option D is correct.
Quadratic equation will become very easy among the students if they are well known about speed, accuracy and very important which is strategy.As a Conclusion, these are the basic concepts of attempting Banking Exams's quadratic equations questions. The main challenge of the Quadratic equations is to check the kind of question presented which allows you to productively approach and solve it. However, that can only come with practice, practice and practice because as we all heard the fact "Practice Makes Man Perfect", and hence, keep practicing Quadratic Equations questions regularly to gives you intelligence among this topic.
Quadratic Equation Quizzes:
- Quadratic Equation for SBI Clerk
- Quadratic Equation Questions for SBI Clerk
- Quadratic Equation for RBI Assistant
- Quadratic Equation Questions for RBI Assistant
- Quadratic Equation Questions for Bank
- Quadratic Equation for Bank Exam
In this article we have covered given below queries:
- New Pattern Quadratic Equation PDF
- Quadratic Equation PDF
- Quadratic Equation PDF for Bank Exam
- Quadratic Equation for SBI Clerk PDF
- Quadratic Equation Questions and Answers for Bank Exam PDF
- Quadratic Equation Questions PDF
- Quadratic Equation Questions for Competitive Exam
- Quadratic Equation Questions and Answers for Bank Exam
- Quadratic Equations PDF
All the Best
Regards
Team Smartkeeda