{"id":101671,"date":"2025-04-09T10:38:18","date_gmt":"2025-04-09T05:08:18","guid":{"rendered":"https:\/\/www.practicemock.com\/blog\/?p=101671"},"modified":"2025-04-17T13:17:26","modified_gmt":"2025-04-17T07:47:26","slug":"quadratic-equation-for-ibps-rrb-exam","status":"publish","type":"post","link":"https:\/\/www.practicemock.com\/blog\/quadratic-equation-for-ibps-rrb-exam\/","title":{"rendered":"Quadratic Equation For IBPS RRB Exam, Check Important Question"},"content":{"rendered":"\n

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Home<\/a><\/span> \u00bb Banking & Insurance<\/a><\/span> \u00bb Quadratic Equation For IBPS RRB Exam<\/span><\/span><\/div>\n\n\n

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Quadratic Equation for IBPS RRB Exam- <\/strong>The topic of quadratic equations is a crucial part of the quantitative aptitude section in various competitive exams, including the IBPS RRB exam and other banking exams. Here we are going to discuss a few Quant Quadratic Equation Practice Questions for the Prelims Exam. These questions will be helpful for the upcoming IBPS RRB 2024 Exam. Though it is one of the trickiest topics from this section, these can be solved with constant practice which is why we are here with these practice questions along with their answers and detailed solutions. Mastering these equations is essential for scoring well in this section. Candidates check all details in the article below.<\/p>\n\n\n\n

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Take a IBPS RRB PO Free Mock Test & Know the Difficulty Level of the Questions Asked<\/strong><\/a><\/div>\n<\/div>\n\n\n\n

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Take 1 Free Mock Test & Measure Your Performance in Numbers<\/strong><\/a><\/div>\n<\/div>\n\n\n\n

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What is a Quadratic Equation?<\/h2>\n\n\n\n

A quadratic equation is a second-degree polynomial equation in one variable of the form:<\/p>\n\n\n\n

ax\u00b2 + bx + c = 0<\/strong><\/p>\n\n\n\n

where a, b and c are constants, and the x is variable. The x represents the unknown quantity, and the constants a, b and c determine the shape and nature of the equation.<\/p>\n\n\n\n

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IBPS RRB Notification 2025- Click To Check<\/strong><\/a><\/div>\n<\/div>\n\n\n\n

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Important Question for Quadratic Equation<\/h2>\n\n\n\n

Question 1:<\/strong> In the question, two equations I and II are given. You have to solve both the equations to establish the correct relation between x and y and choose the correct option.<\/p>\n\n\n\n

I. 3x2<\/sup> \u2013 24x + 36 = 0<\/p>\n\n\n\n

II. 4y2<\/sup> = y + 5<\/p>\n\n\n\n

A) x > y<\/p>\n\n\n\n

B) x < y<\/p>\n\n\n\n

C) x = y or the relationship cannot be established<\/p>\n\n\n\n

D) x \u2265 y<\/p>\n\n\n\n

E) x \u2264 y<\/p>\n\n\n\n

Question 2:<\/strong> In the question, two equations I and II are given. You have to solve both the equations to establish the correct relation between x and y and choose the correct option.<\/p>\n\n\n\n

I. x2<\/sup> \u2013 2x \u2013 63 = 0<\/p>\n\n\n\n

II. y2<\/sup> + 14y + 48 = 0<\/p>\n\n\n\n

A) x > y<\/p>\n\n\n\n

B) x < y<\/p>\n\n\n\n

C) x = y or the relationship cannot be established<\/p>\n\n\n\n

D) x \u2265 y<\/p>\n\n\n\n

E) x \u2264 y<\/p>\n\n\n\n

Question 3:<\/strong> In the question, two equations I and II are given. You have to solve both the equations to establish the correct relation between x and y and choose the correct option.<\/p>\n\n\n\n

I. x2<\/sup> + 15x + 54 = 0<\/p>\n\n\n\n

II. y2<\/sup> + 20y + 99 = 0<\/p>\n\n\n\n

A) x > y<\/p>\n\n\n\n

B) x < y<\/p>\n\n\n\n

C) x = y or the relationship cannot be established<\/p>\n\n\n\n

D) x \u2265 y<\/p>\n\n\n\n

E) x \u2264 y<\/p>\n\n\n\n

Question 4:<\/strong> In the question, two equations I and II are given. You have to solve both the equations to establish the correct relation between x and y and choose the correct option.<\/p>\n\n\n\n

I. 3x + 7y = 53<\/p>\n\n\n\n

II. 7x \u2013 5y = 17<\/p>\n\n\n\n

A) x > y<\/p>\n\n\n\n

B) x < y<\/p>\n\n\n\n

C) x = y or the relationship cannot be established<\/p>\n\n\n\n

D) x \u2265 y<\/p>\n\n\n\n

E) x \u2264 y<\/p>\n\n\n\n

Question 5:<\/strong> In the question, two equations I and II are given. You have to solve both the equations to establish the correct relation between x and y and choose the correct option.<\/p>\n\n\n\n

I. x2<\/sup> \u2013 27x + 180 = 0<\/p>\n\n\n\n

II. y2<\/sup> \u2013 24y + 140 = 0<\/p>\n\n\n\n

A) x > y<\/p>\n\n\n\n

B) x < y<\/p>\n\n\n\n

C) x = y or the relationship cannot be established<\/p>\n\n\n\n

D) x \u2265 y<\/p>\n\n\n\n

E) x \u2264 y<\/p>\n\n\n\n

Question 6:<\/strong> In the question, two equations I and II are given. You have to solve both the equations to establish the correct relation between x and y and choose the correct option.<\/p>\n\n\n\n

I. x2<\/sup> \u2013 11x + 30 = 0<\/p>\n\n\n\n

II. y2<\/sup> \u2013 11y + 18 = 0<\/p>\n\n\n\n

A) x > y<\/p>\n\n\n\n

B) x < y<\/p>\n\n\n\n

C) x = y or the relationship cannot be established<\/p>\n\n\n\n

D) x \u2265 y<\/p>\n\n\n\n

E) x \u2264 y<\/p>\n\n\n\n

Question 7:<\/strong> In the question, two equations I and II are given. You have to solve both the equations to establish the correct relation between x and y and choose the correct option.<\/p>\n\n\n\n

I. x2<\/sup> + 3x \u2013 10 = 0<\/p>\n\n\n\n

II. y2 <\/sup>\u2013 5y + 6 = 0<\/p>\n\n\n\n

A) x > y<\/p>\n\n\n\n

B) x < y<\/p>\n\n\n\n

C) x = y or the relationship cannot be established<\/p>\n\n\n\n

D) x \u2265 y<\/p>\n\n\n\n

E) x \u2264 y<\/p>\n\n\n\n

Question 8:<\/strong> In the question, two equations I and II are given. You have to solve both the equations to establish the correct relation between x and y and choose the correct option.<\/p>\n\n\n\n

I. x2<\/sup> \u2013 5x + 4 = 0<\/p>\n\n\n\n

II. y2 <\/sup>+ 2y \u2013 3 = 0<\/p>\n\n\n\n

A) x > y<\/p>\n\n\n\n

B) x < y<\/p>\n\n\n\n

C) x = y or the relationship cannot be established<\/p>\n\n\n\n

D) x \u2265 y<\/p>\n\n\n\n

E) x \u2264 y<\/p>\n\n\n\n

Question 9:<\/strong> In the question, two equations I and II are given. You have to solve both the equations to establish the correct relation between x and y and choose the correct option.<\/p>\n\n\n\n

I. 3x + 5y = 21<\/p>\n\n\n\n

II. 9x \u2013 2y = 12<\/p>\n\n\n\n

A) x > y<\/p>\n\n\n\n

B) x < y<\/p>\n\n\n\n

C) x = y or the relationship cannot be established<\/p>\n\n\n\n

D) x \u2265 y<\/p>\n\n\n\n

E) x \u2264 y<\/p>\n\n\n\n

Question 10:<\/strong> In the question, two equations I and II are given. You have to solve both the equations to establish the correct relation between x and y and choose the correct option.<\/p>\n\n\n\n

I. 4x2<\/sup> + 16x \u2013 20 = 0<\/p>\n\n\n\n

II. 2y2<\/sup> + 11y \u2013 6 = 0<\/p>\n\n\n\n

A) x > y<\/p>\n\n\n\n

B) x < y<\/p>\n\n\n\n

C) x = y or the relationship cannot be established<\/p>\n\n\n\n

D) x \u2265 y<\/p>\n\n\n\n

E) x \u2264 y<\/p>\n\n\n\n

Solution 1: A)<\/strong><\/p>\n\n\n\n

From I:<\/p>\n\n\n\n

3x2<\/sup> \u2013 24x + 36 = 0<\/p>\n\n\n\n

3x2<\/sup> \u2013 18x \u2013 6x + 36 = 0<\/p>\n\n\n\n

3x(x \u2013 6) \u2013 6(x \u2013 6) = 0<\/p>\n\n\n\n

(3x \u2013 6)(x \u2013 6) = 0<\/p>\n\n\n\n

x = 2, 6<\/p>\n\n\n\n

From II:<\/p>\n\n\n\n

4y2<\/sup> = y + 5<\/p>\n\n\n\n

4y2<\/sup> \u2013 y \u2013 5 = 0<\/p>\n\n\n\n

4y2<\/sup> \u2013 5y + 4y \u2013 5 = 0<\/p>\n\n\n\n

y(4y \u2013 5) + 1(4y \u2013 5) = 0<\/p>\n\n\n\n

(4y \u2013 5)(y + 1) = 0<\/p>\n\n\n\n

y = -1, 5\/4<\/p>\n\n\n\n

So x > y<\/p>\n\n\n\n

Hence, option a.<\/p>\n\n\n\n

Solution 2: C)<\/strong><\/p>\n\n\n\n

From I:<\/p>\n\n\n\n

x2<\/sup> \u2013 2x \u2013 63 = 0<\/p>\n\n\n\n

x2<\/sup> \u2013 9x + 7x \u2013 63 = 0<\/p>\n\n\n\n

x(x \u2013 9) + 7(x \u2013 9) = 0<\/p>\n\n\n\n

(x + 7)(x \u2013 9) = 0<\/p>\n\n\n\n

x = 9, -7<\/p>\n\n\n\n

From II:<\/p>\n\n\n\n

y2<\/sup> + 14y + 48 = 0<\/p>\n\n\n\n

y2<\/sup> + 8y + 6y + 48 = 0<\/p>\n\n\n\n

y(y + 8) + 6(y + 8) = 0<\/p>\n\n\n\n

(y + 8)(y + 6) = 0<\/p>\n\n\n\n

y = -6, -8<\/p>\n\n\n\n

No relation can be established between x and y.<\/p>\n\n\n\n

Hence, option c.<\/p>\n\n\n\n

Solution 3: D)<\/strong><\/p>\n\n\n\n

From I:<\/p>\n\n\n\n

x2<\/sup> + 15x + 54 = 0<\/p>\n\n\n\n

x2<\/sup> + 9x + 6x + 54 = 0<\/p>\n\n\n\n

x(x + 9) + 6(x + 9) = 0<\/p>\n\n\n\n

(x + 6)(x + 9) = 0<\/p>\n\n\n\n

x = -6, -9<\/p>\n\n\n\n

From II:<\/p>\n\n\n\n

y2<\/sup> + 20y + 99 = 0<\/p>\n\n\n\n

y2<\/sup> + 11y + 9y + 99 = 0<\/p>\n\n\n\n

y(y + 11) + 9(y + 11) = 0<\/p>\n\n\n\n

(y + 11)(y + 9) = 0<\/p>\n\n\n\n

y = -9, -11<\/p>\n\n\n\n

So x \u2265 y<\/p>\n\n\n\n

Hence, option d.<\/p>\n\n\n\n

Solution 4: A)<\/strong><\/p>\n\n\n\n

From I:<\/p>\n\n\n\n

3x + 7y = 53<\/p>\n\n\n\n

3x = 53 \u2013 7y<\/p>\n\n\n\n

x = (53 \u2013 7y)\/3<\/p>\n\n\n\n

From II:<\/p>\n\n\n\n

7x \u2013 5y = 17<\/p>\n\n\n\n

7 \u00d7 {(53 \u2013 7y)\/3} \u2013 5y = 17<\/p>\n\n\n\n

371 \u2013 49y \u2013 15y = 51<\/p>\n\n\n\n

64y = 320<\/p>\n\n\n\n

y = 5<\/p>\n\n\n\n

Now x = {53 \u2013 7 \u00d7 5}\/3<\/p>\n\n\n\n

x = 18\/3<\/p>\n\n\n\n

x = 6<\/p>\n\n\n\n

So x > y<\/p>\n\n\n\n

Hence, option a.<\/p>\n\n\n\n

Solution 5: C)<\/strong><\/p>\n\n\n\n

From I:<\/p>\n\n\n\n

x2<\/sup> \u2013 27x + 180 = 0<\/p>\n\n\n\n

x2<\/sup> \u2013 15x  \u2013 12x + 180 = 0<\/p>\n\n\n\n

x(x \u2013 15) \u2013 12(x \u2013 15) = 0<\/p>\n\n\n\n

(x \u2013 15)(x \u2013 12) = 0<\/p>\n\n\n\n

x = 15, 12<\/p>\n\n\n\n

From II:<\/p>\n\n\n\n

y2<\/sup> \u2013 24y + 140 = 0<\/p>\n\n\n\n

y2<\/sup> \u2013 14y \u2013 10y + 140 = 0<\/p>\n\n\n\n

y(y \u2013 14) \u2013 10(y \u2013 14) = 0<\/p>\n\n\n\n

(y \u2013 10)(y \u2013 14) = 0<\/p>\n\n\n\n

y = 14, 10<\/p>\n\n\n\n

No relation can be established between x and y.<\/p>\n\n\n\n

Hence, option c.<\/p>\n\n\n\n

Solution 6: C)<\/strong><\/p>\n\n\n\n

From I:<\/p>\n\n\n\n

x2<\/sup> \u2013 11x + 30 = 0<\/p>\n\n\n\n

x2<\/sup> – 6x \u2013 5x + 30 = 0<\/p>\n\n\n\n

x(x \u2013 6) \u2013 5(x – 6) = 0<\/p>\n\n\n\n

(x – 6)(x – 5) = 0<\/p>\n\n\n\n

x = 6, 5<\/p>\n\n\n\n

From II:<\/p>\n\n\n\n

y2<\/sup> \u2013 11y + 18 = 0<\/p>\n\n\n\n

y2<\/sup> \u2013 2y \u2013 9y + 18 = 0<\/p>\n\n\n\n

y(y \u2013 2) \u2013 9(y \u2013 2) = 0<\/p>\n\n\n\n

(y \u2013 2)(y \u2013 9) = 0<\/p>\n\n\n\n

y = 2, 9<\/p>\n\n\n\n

Hence, option c.<\/p>\n\n\n\n

Solution 7: E)<\/strong><\/p>\n\n\n\n

From I:<\/p>\n\n\n\n

x2<\/sup> + 3x \u2013 10 = 0<\/p>\n\n\n\n

x2<\/sup> \u2013 2x + 5x \u2013 10 = 0<\/p>\n\n\n\n

x(x \u2013 2) + 5(x \u2013 2) = 0<\/p>\n\n\n\n

(x \u2013 2)(x + 5) = 0<\/p>\n\n\n\n

x = 2, -5<\/p>\n\n\n\n

From II:<\/p>\n\n\n\n

y2 <\/sup>\u2013 5y + 6 = 0<\/p>\n\n\n\n

y2 <\/sup>\u2013 3y \u2013 2y + 6 = 0                        <\/p>\n\n\n\n

y(y \u2013 3) \u2013 2(y \u2013 3) = 0<\/p>\n\n\n\n

(y \u2013 3)(y \u2013 2) = 0<\/p>\n\n\n\n

y = 3, 2<\/p>\n\n\n\n

So, x \u2264 y<\/p>\n\n\n\n

Hence, option e.<\/p>\n\n\n\n

Solution 8: D)<\/strong><\/p>\n\n\n\n

From I:<\/p>\n\n\n\n

x2<\/sup> \u2013 5x + 4 = 0<\/p>\n\n\n\n

x2<\/sup> \u2013 x \u2013 4x + 4 = 0<\/p>\n\n\n\n

x(x \u2013 1) \u2013 4(x \u2013 1) = 0<\/p>\n\n\n\n

(x \u2013 1)(x \u2013 4) = 0<\/p>\n\n\n\n

x = 1, 4<\/p>\n\n\n\n

From II:<\/p>\n\n\n\n

y2 <\/sup>+ 2y \u2013 3 = 0<\/p>\n\n\n\n

y2<\/sup> + 3y \u2013 y \u2013 3 = 0<\/p>\n\n\n\n

y(y + 3) \u2013 1(y + 3) = 0<\/p>\n\n\n\n

(y + 3)(y \u2013 1) = 0<\/p>\n\n\n\n

y = -3, 1<\/p>\n\n\n\n

Hence, option d.<\/p>\n\n\n\n

Solution 9: B)<\/strong><\/p>\n\n\n\n

From I:<\/p>\n\n\n\n

3x + 5y = 21<\/p>\n\n\n\n

3x = 21 \u2013 5y<\/p>\n\n\n\n

x = (21 \u2013 5y)\/3<\/p>\n\n\n\n

From II:<\/p>\n\n\n\n

9x \u2013 2y = 12<\/p>\n\n\n\n

9 \u00d7 {(21 \u2013 5y)\/3} \u2013 2y = 12<\/p>\n\n\n\n

(189 \u2013 45y \u2013 6y)\/3 = 12              <\/p>\n\n\n\n

189 \u2013 51y = 36<\/p>\n\n\n\n

51y = 153, y = 3<\/p>\n\n\n\n

x = {(21 \u2013 5y)\/3} = 6\/3 = 2<\/p>\n\n\n\n

So, x < y<\/p>\n\n\n\n

Hence, option b.<\/p>\n\n\n\n

Solution 10: C)<\/strong><\/p>\n\n\n\n

From I:<\/p>\n\n\n\n

4x2<\/sup> + 16x \u2013 20 = 0<\/p>\n\n\n\n

4x2<\/sup> + 20x \u2013 4x \u2013 20 = 0<\/p>\n\n\n\n

x (4x + 20) \u2013 1(4x + 20) = 0<\/p>\n\n\n\n

(4x + 20)(x \u2013 1) = 0<\/p>\n\n\n\n

x = -5, 1<\/p>\n\n\n\n

From II:<\/p>\n\n\n\n

2y2<\/sup> + 11y \u2013 6 = 0<\/p>\n\n\n\n

2y2<\/sup> + 12y \u2013 y \u2013 6 = 0<\/p>\n\n\n\n

2y(y + 6) \u2013 1(y + 6) = 0<\/p>\n\n\n\n

(y + 6)(2y \u2013 1) = 0<\/p>\n\n\n\n

y = \u2013 6, 1\/2<\/p>\n\n\n\n

Hence, option c.<\/p>\n\n\n\n

Related Blogs<\/h3>\n\n\n\n

For More Details Regarding the IBPS RRB 2025 Notification for PO and Clerk, candidates can click on the provided links. Important details of the IBPS RRB PO and Clerk, such as syllabus, exam pattern, cut-off, online application, and previous year papers, are provided. <\/p>\n\n\n\n

IBPS RRB Clerk Other Article<\/h3>\n\n\n\n
IBPS RRB Clerk Exam Pattern 2025<\/a><\/td>IBPS RRB Clerk Syllabus 2025<\/a><\/td><\/tr>
IBPS RRB Clerk Salary 2025<\/a><\/td>IBPS RRB Clerk Cut Off 2025<\/a><\/td><\/tr>
IBPS RRB Clerk Previous Year Paper<\/a><\/td><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

IBPS RRB PO Other Article<\/h3>\n\n\n\n
IBPS RRB PO Salary<\/a><\/td>IBPS RRB PO Exam Pattern<\/a><\/td><\/tr>
IBPS RRB PO Cut Off<\/a><\/td>IBPS RRB PO Syllabus<\/a><\/td><\/tr>
IBPS RRB PO Previous Year Question Papers<\/a><\/td><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

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Join our exclusive Telegram group where our experts are ready to answer all your queries, guide you in banking exam preparation, and give personalized tips to boost your success. Get access to real-time solutions, expert advice, and valuable resources to improve your study journey. [Click here to join now!]<\/strong><\/em><\/p>\n\n\n\n

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PracticeMock Telegram group Link<\/strong><\/a><\/div>\n<\/div>\n\n\n\n

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\"Banking<\/a>
<\/figcaption><\/figure>\n\n\n\n

Quadratic Equation For IBPS RRB Exam FAQ<\/h2>\n\n\n\n
What is a quadratic equation?<\/strong><\/strong>

A quadratic equation is a polynomial equation of the form ax2+bx+c=0ax^2 + bx + c = 0ax2+bx+c=0, where a, b and c are constants, and x is a variable.<\/p> <\/div>

What are the methods to solve a quadratic equation?<\/strong><\/strong>

The methods to solve a quadratic equation include: Factoring, Using the quadratic formula, Completing the square, and Graphing<\/p> <\/div>

<\/strong>

<\/p> <\/div> <\/div>\n","protected":false},"excerpt":{"rendered":"

Candidates Boost your preparation for Quadratic Equation For IBPS RRB Exam with our tips and trick and important questions.<\/p>\n","protected":false},"author":13,"featured_media":101672,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_uag_custom_page_level_css":"","_monsterinsights_skip_tracking":false,"_monsterinsights_sitenote_active":false,"_monsterinsights_sitenote_note":"","_monsterinsights_sitenote_category":0,"_uf_show_specific_survey":0,"_uf_disable_surveys":false,"footnotes":""},"categories":[22],"tags":[],"class_list":["post-101671","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-banking-insurance"],"yoast_head":"\nQuadratic Equation For IBPS RRB Exam, Check Important Question<\/title>\n<meta name=\"description\" content=\"Candidates Boost your preparation for Quadratic Equation For IBPS RRB Exam with our tips and trick and important questions.\" \/>\n<meta name=\"robots\" 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I cleared many bank exams time by time but couldn't join because of my passion towards writing. I write blogs to help aspirants prepare for Banking and Insurance exams. These blogs turn out to be a one-stop destination for comprehensive information on some of the biggest competitive exams like SBI PO\/Clerk, IBPS PO\/Clerk, IBPS RRB PO\/Clerk and RBI. My ultimate goal is to provide accurate and easy-to-understand information, covering topics like exam patterns, syllabus, study techniques, and more. Join me on this journey of knowledge!","url":"https:\/\/www.practicemock.com\/blog\/author\/ss8860409gmail-com\/"},{"@type":"Question","@id":"https:\/\/www.practicemock.com\/blog\/quadratic-equation-for-ibps-rrb-exam\/#faq-question-1721039976270","position":1,"url":"https:\/\/www.practicemock.com\/blog\/quadratic-equation-for-ibps-rrb-exam\/#faq-question-1721039976270","name":"What is a quadratic equation?","answerCount":1,"acceptedAnswer":{"@type":"Answer","text":"A quadratic equation is a polynomial equation of the form ax2+bx+c=0ax^2 + bx + c = 0ax2+bx+c=0, where a, b and c are constants, and x is a variable.","inLanguage":"en-US"},"inLanguage":"en-US"},{"@type":"Question","@id":"https:\/\/www.practicemock.com\/blog\/quadratic-equation-for-ibps-rrb-exam\/#faq-question-1721040261324","position":2,"url":"https:\/\/www.practicemock.com\/blog\/quadratic-equation-for-ibps-rrb-exam\/#faq-question-1721040261324","name":"What are the methods to solve a quadratic equation?","answerCount":1,"acceptedAnswer":{"@type":"Answer","text":"The methods to solve a quadratic equation include: Factoring, Using the quadratic formula, Completing the square, and Graphing","inLanguage":"en-US"},"inLanguage":"en-US"}]}},"uagb_featured_image_src":{"full":["https:\/\/www.practicemock.com\/blog\/wp-content\/uploads\/2024\/07\/Quadratic-Equation-For-IBPS-RRB-Exam-1.png",1280,768,false],"thumbnail":["https:\/\/www.practicemock.com\/blog\/wp-content\/uploads\/2024\/07\/Quadratic-Equation-For-IBPS-RRB-Exam-1.png",150,90,false],"medium":["https:\/\/www.practicemock.com\/blog\/wp-content\/uploads\/2024\/07\/Quadratic-Equation-For-IBPS-RRB-Exam-1.png",300,180,false],"medium_large":["https:\/\/www.practicemock.com\/blog\/wp-content\/uploads\/2024\/07\/Quadratic-Equation-For-IBPS-RRB-Exam-1.png",640,384,false],"large":["https:\/\/www.practicemock.com\/blog\/wp-content\/uploads\/2024\/07\/Quadratic-Equation-For-IBPS-RRB-Exam-1.png",640,384,false],"1536x1536":["https:\/\/www.practicemock.com\/blog\/wp-content\/uploads\/2024\/07\/Quadratic-Equation-For-IBPS-RRB-Exam-1.png",1280,768,false],"2048x2048":["https:\/\/www.practicemock.com\/blog\/wp-content\/uploads\/2024\/07\/Quadratic-Equation-For-IBPS-RRB-Exam-1.png",1280,768,false],"web-stories-poster-portrait":["https:\/\/www.practicemock.com\/blog\/wp-content\/uploads\/2024\/07\/Quadratic-Equation-For-IBPS-RRB-Exam-1.png",640,384,false],"web-stories-publisher-logo":["https:\/\/www.practicemock.com\/blog\/wp-content\/uploads\/2024\/07\/Quadratic-Equation-For-IBPS-RRB-Exam-1.png",96,58,false],"web-stories-thumbnail":["https:\/\/www.practicemock.com\/blog\/wp-content\/uploads\/2024\/07\/Quadratic-Equation-For-IBPS-RRB-Exam-1.png",150,90,false]},"uagb_author_info":{"display_name":"Sweta Singh","author_link":"https:\/\/www.practicemock.com\/blog\/author\/ss8860409gmail-com\/"},"uagb_comment_info":32,"uagb_excerpt":"Candidates Boost your preparation for Quadratic Equation For IBPS RRB Exam with our tips and trick and important questions.","_links":{"self":[{"href":"https:\/\/www.practicemock.com\/blog\/wp-json\/wp\/v2\/posts\/101671","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.practicemock.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.practicemock.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.practicemock.com\/blog\/wp-json\/wp\/v2\/users\/13"}],"replies":[{"embeddable":true,"href":"https:\/\/www.practicemock.com\/blog\/wp-json\/wp\/v2\/comments?post=101671"}],"version-history":[{"count":0,"href":"https:\/\/www.practicemock.com\/blog\/wp-json\/wp\/v2\/posts\/101671\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.practicemock.com\/blog\/wp-json\/wp\/v2\/media\/101672"}],"wp:attachment":[{"href":"https:\/\/www.practicemock.com\/blog\/wp-json\/wp\/v2\/media?parent=101671"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.practicemock.com\/blog\/wp-json\/wp\/v2\/categories?post=101671"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.practicemock.com\/blog\/wp-json\/wp\/v2\/tags?post=101671"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}