{"id":10662,"date":"2021-11-08T11:20:43","date_gmt":"2021-11-08T05:50:43","guid":{"rendered":"https:\/\/www.practicemock.com\/blog\/?p=10662"},"modified":"2021-11-08T11:20:43","modified_gmt":"2021-11-08T05:50:43","slug":"quadratic-equation-for-sbi-po","status":"publish","type":"post","link":"https:\/\/www.practicemock.com\/blog\/quadratic-equation-for-sbi-po\/","title":{"rendered":"Quadratic Equation for SBI PO Questions"},"content":{"rendered":"\n

Quadratic Equation for SBI PO: The questions from the quadratic equation are more or less tricky and difficult in nature. But by practicing the right set of questions they are as easy as any other quant topic. We today have six questions, 3 each of prelims and mains with answer key. <\/p>\n\n\n\n

\"Quadratic<\/a><\/figure>\n\n\n\n

Quadratic Equation for SBI PO Pr<\/strong>elims Questions<\/h2>\n\n\n\n

1.) 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. (x + 4)2<\/sup> = 23x \u2013 20<\/p>\n\n\n\n

II. (y \u2013 3)2<\/sup> = 3 \u2013 y<\/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

Answer: d)<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

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

(x + 4)2<\/sup> = 23x \u2013 20<\/p>\n\n\n\n

x2<\/sup> + 8x + 16 = 23x \u2013 20<\/p>\n\n\n\n

x2<\/sup> \u2013 15x + 36 = 0<\/p>\n\n\n\n

x2<\/sup> \u2013 3x \u2013 12x + 36 = 0<\/p>\n\n\n\n

x(x \u2013 3) \u2013 12(x \u2013 3) = 0<\/p>\n\n\n\n

(x \u2013 3)(x \u2013 12) = 0<\/p>\n\n\n\n

x = 12, 3<\/p>\n\n\n\n

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

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

y2<\/sup> \u2013 6y + 9 = 3 \u2013 y<\/p>\n\n\n\n

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

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

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

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

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

X<\/td>Relation<\/td>y<\/td><\/tr>
12<\/td>><\/td>2<\/td><\/tr>
12<\/td>><\/td>3<\/td><\/tr>
3<\/td>><\/td>2<\/td><\/tr>
3<\/td>=<\/td>3<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

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

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

\"SBI_PO_Free<\/a><\/figure>\n\n\n\n

2.) 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> = 324<\/p>\n\n\n\n

II. y2 <\/sup>+ 18y = 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

Answer: c)<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

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

x2<\/sup> = 324<\/p>\n\n\n\n

x = -18, 18<\/p>\n\n\n\n

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

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

y(y + 18) = 0<\/p>\n\n\n\n

y = 0, -18<\/p>\n\n\n\n

X<\/td>Relation<\/td>y<\/td><\/tr>
-18<\/td><<\/td>0<\/td><\/tr>
-18<\/td>=<\/td>-18<\/td><\/tr>
18<\/td>><\/td>0<\/td><\/tr>
18<\/td>><\/td>-18<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

So, the relationship cannot be established.<\/p>\n\n\n\n

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

3.) 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. x3<\/sup> = -729<\/p>\n\n\n\n

II. y2 <\/sup>+ 15y + 54 = 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

Answer: e)<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

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

x3<\/sup> = -729<\/p>\n\n\n\n

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

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

y2 <\/sup>+ 15y + 54 = 0<\/p>\n\n\n\n

y2<\/sup> + 6y + 9y + 54 = 0<\/p>\n\n\n\n

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

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

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

X<\/td>Relation<\/td>y<\/td><\/tr>
-9<\/td>=<\/td>-9<\/td><\/tr>
-9<\/td><<\/td>-6<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

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

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

\"Quadratic<\/a><\/figure>\n\n\n\n

Quadratic Equation for SBI PO Mains Questions<\/h2>\n\n\n\n

1.) 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. x + 19\u221ax + 90 = 0<\/p>\n\n\n\n

II. y + 23\u221ay + 130 = 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

Answer: e)<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

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

x + 19\u221ax + 90 = 0<\/p>\n\n\n\n

x + 9\u221ax + 10\u221ax + 90 = 0<\/p>\n\n\n\n

\u221ax(\u221ax + 9) + 10(\u221ax + 9) = 0<\/p>\n\n\n\n

(\u221ax + 10)(\u221ax + 9) = 0<\/p>\n\n\n\n

\u221ax = -10, -9<\/p>\n\n\n\n

x = 100, 81<\/p>\n\n\n\n

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

y + 23\u221ay + 130 = 0<\/p>\n\n\n\n

y + 10\u221ay + 13\u221ay + 130 = 0<\/p>\n\n\n\n

\u221ay(\u221ay + 10) + 13(\u221ay + 10) = 0<\/p>\n\n\n\n

(\u221ay + 10)(\u221ay + 13) = 0<\/p>\n\n\n\n

\u221ay = -10, -13<\/p>\n\n\n\n

y = 100, 169<\/p>\n\n\n\n

x<\/td>Relation<\/td>y<\/td><\/tr>
100<\/td>=<\/td>100<\/td><\/tr>
100<\/td><<\/td>169<\/td><\/tr>
81<\/td><<\/td>100<\/td><\/tr>
81<\/td><<\/td>169<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

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

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

2.) 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> + (9\u221a3)x + 60 = 0<\/p>\n\n\n\n

II. y2<\/sup> + (6 + 4\u221a3)y + 24\u221a3 = 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

Answer: e)<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

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

x2<\/sup> + (9\u221a3)x + 60 = 0<\/p>\n\n\n\n

x2<\/sup> + (5\u221a3)x + (4\u221a3)x + 60 = 0<\/p>\n\n\n\n

x(x + 5\u221a3) + 4\u221a3(x + 5\u221a3) = 0<\/p>\n\n\n\n

(x + 5\u221a3)(x + 4\u221a3) = 0<\/p>\n\n\n\n

x = -5\u221a3, -4\u221a3<\/p>\n\n\n\n

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

y2<\/sup> + (6 + 4\u221a3)y + 24\u221a3 = 0<\/p>\n\n\n\n

y2<\/sup> + 6y + (4\u221a3)y + 24\u221a3 = 0<\/p>\n\n\n\n

y(y + 6) + 4\u221a3(y + 6) = 0<\/p>\n\n\n\n

(y + 4\u221a3)(y + 6) = 0<\/p>\n\n\n\n

y = -4\u221a3, -6<\/p>\n\n\n\n

x<\/td>Relation<\/td>y<\/td><\/tr>
-5\u221a3<\/td><<\/td>-4\u221a3<\/td><\/tr>
-5\u221a3<\/td><<\/td>-6<\/td><\/tr>
-4\u221a3<\/td>=<\/td>-4\u221a3<\/td><\/tr>
-4\u221a3<\/td><<\/td>-6<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

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

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

\"SBI_PO_Free<\/a><\/figure>\n\n\n\n

3.) 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. 81x2<\/sup>+ 144x + 55 = 0<\/p>\n\n\n\n

II. 81y2<\/sup>+ 180y + 91 = 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

Answer: c)<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

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

81x2<\/sup>+ 144x + 55 = 0<\/p>\n\n\n\n

81x2<\/sup>+ 99x + 45x + 55 = 0<\/p>\n\n\n\n

9x(9x + 11) + 5(9x + 11) = 0<\/p>\n\n\n\n

(9x + 5)(9x + 11) = 0<\/p>\n\n\n\n

x = -5\/9, -11\/9<\/p>\n\n\n\n

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

81y2<\/sup>+ 180y + 91 = 0<\/p>\n\n\n\n

81y2<\/sup>+ 117y + 63y + 91 = 0<\/p>\n\n\n\n

9y(9y + 13) + 7(9y + 13) = 0<\/p>\n\n\n\n

(9y + 7)(9y + 13) = 0<\/p>\n\n\n\n

y = -7\/9, -13\/9<\/p>\n\n\n\n

x<\/td>Relation<\/td>y<\/td><\/tr>
-5\/9<\/td>><\/td>-7\/9<\/td><\/tr>
-5\/9<\/td>><\/td>-13\/9<\/td><\/tr>
-11\/9<\/td><<\/td>-7\/9<\/td><\/tr>
-11\/9<\/td>><\/td>-13\/9<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

So, the relationship cannot be established<\/p>\n\n\n\n

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

\"Quadratic<\/a><\/figure>\n\n\n\n

Find daily exam updates posted on our official Instagram handle <\/a>as well.   <\/p>\n\n\n\n

You might want to subscribe to our push notifications for such daily valuable articles.            <\/p>\n\n\n\n

Still, have some doubts about what we discussed in this blog? Then, leave your queries in the comment section below. We will surely respond to them. <\/p>\n","protected":false},"excerpt":{"rendered":"

The Quadratic Equation for SBI PO questions for both prelims and mains practice are given. Read the article for six SBI PO oriented questions.<\/p>\n","protected":false},"author":5,"featured_media":10664,"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":[49],"tags":[1151,1152,427,230,1026,1027],"class_list":["post-10662","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-sbi-po","tag-quadratic-equations","tag-quadratic-questions","tag-sbi-po-2021","tag-sbi-po-exam-preparation","tag-sbi-po-quant","tag-sbi-po-quantitative-aptitude"],"yoast_head":"\nQuadratic Equation for SBI PO Questions - Practicemock<\/title>\n<meta name=\"description\" content=\"The Quadratic Equation for SBI PO questions for both prelims and mains practice are given. 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