{"id":4365,"date":"2023-02-09T13:51:32","date_gmt":"2023-02-09T08:21:32","guid":{"rendered":"https:\/\/apboardsolutions.com\/?p=4365"},"modified":"2023-03-29T16:07:52","modified_gmt":"2023-03-29T10:37:52","slug":"ap-ssc-10th-class-maths-solutions-chapter-5-ex-5-1","status":"publish","type":"post","link":"https:\/\/apboardsolutions.com\/ap-ssc-10th-class-maths-solutions-chapter-5-ex-5-1\/","title":{"rendered":"AP SSC 10th Class Maths Solutions Chapter 5 Quadratic Equations Ex 5.1"},"content":{"rendered":"

AP State Board Syllabus\u00a0AP SSC 10th Class Maths Textbook Solutions<\/a> Chapter 5 Quadratic Equations Ex 5.1 Textbook Questions and Answers.<\/p>\n

AP State Syllabus SSC 10th Class Maths Solutions 5th Lesson Quadratic Equations Exercise 5.1<\/h2>\n

10th Class Maths 5th Lesson Quadratic Equations Ex 5.1 Textbook Questions and Answers<\/h3>\n

Question 1.
\nCheck whether the following are quadratic equations.
\ni) (x + l)2<\/sup> = 2(x-3)
\nAnswer:
\nGiven: (x + l)2<\/sup> = 2(x – 3)
\n\u21d2 x2<\/sup> + 2x + 1 = 2(x – 3) = 2x – 6
\n\u21d2 x2<\/sup> + 2x + l – 2x + 6 = 0
\n\u21d2 x2<\/sup> + 7 = 0 is a Q.E.<\/p>\n

ii) x2<\/sup> – 2x = (-2) (3 – x)
\nAnswer:
\nGiven: x2<\/sup> – 2x = -2(3 – x)
\n\u21d2 x2<\/sup> – 2x = -6 + 2x
\n\u21d2 x2<\/sup> – 4x + 6 = 0 is a Q.E.<\/p>\n

\"AP<\/p>\n

iii) (x-2) (x + 1) = (x- 1) (x + 3)
\nAnswer:
\nGiven: (x – 2) (x + 1) = (x – 1) (x + 3)
\n\u21d2 x (x + 1) – 2 (x +1)
\n= x (x + 3) – 1 (x + 3)
\nNote : Compare the coefficients of x2 on both sides. If they are equal it is not a Q.E.
\n\u21d2 x2<\/sup> + x – 2x – 2 = x2<\/sup> + 3x – x -3
\n\u21d2 x2<\/sup> – x – 2 = x2<\/sup> + 2x – 3
\n\u21d2 3x – 1 = 0 is not a Q.E.<\/p>\n

iv) (x – 3) (2x + 1) = x(x + 5)
\nAnswer:
\nGiven: (x – 3) (2x + 1) = x(x + 5)
\n\u21d2 x (2x + 1) – 3 (2x + 1) = x . x + 5 . x
\n\u21d2 2x2<\/sup> + x – 6x – 3 = x2<\/sup> + 5x
\n\u21d2 2x2<\/sup> – 5x – 3 – x2<\/sup> – 5x = 0
\n\u21d2 x2<\/sup> – 10x – 3 = 0 is a Q.E.
\n(or)
\nComparing the coefficients of x2 on both sides.
\nx . 2x and x . x
\n\u21d2 2x2<\/sup> and x2<\/sup>
\n2x2<\/sup> \u2260 x2<\/sup>
\nHence it’s a Q.E.<\/p>\n

v) (2x – 1) (x – 3) = (x + 5) (x – 1)
\nAnswer:
\nGiven: (2x – 1) (x – 3) = (x + 5) (x – 1)
\n\u21d2 2x (x – 3) -1 (x – 3) = x (x – 1) + 5(x – 1)
\n\u21d2 2x2<\/sup> – 6x – x + 3 = x2<\/sup> – x + 5x – 5
\n\u21d2 2x2<\/sup> -7x + 3 – x2<\/sup> – 4x + 5 = 0
\n\u21d2 x2<\/sup> – 11x + 8 = 0
\nHence it’s a Q.E.
\n(or)
\nCo.eff. of x2<\/sup> on L.H.S. = 2 \u00d7 1 = 2
\nCo.eff. of x2<\/sup> on R.H.S = 1 \u00d7 1 = 1
\nLHS \u2260 RHS Hence it is a Q.E.<\/p>\n

vi) x2<\/sup> + 3x + 1 = (x – 2)2<\/sup>
\nAnswer:
\nGiven: x2<\/sup> + 3x + 1 = (x – 2)2<\/sup>
\n\u21d2 x2<\/sup> + 3x + 1 = x2<\/sup> – 4x + 4
\n\u21d2 7x – 3 = 0 is not a Q.E.<\/p>\n

\"AP<\/p>\n

vii) (x + 2)3<\/sup> = 2x (x2<\/sup> – 1)
\nAnswer:
\nGiven: (x + 2)3<\/sup> = 2x(x2<\/sup> – 1)
\n\u21d2 x3<\/sup> + 6x2<\/sup> + 12x + 8 = 2x3<\/sup> – 2x [\u2235 (a + b)3<\/sup> = a3<\/sup> + 3a2<\/sup>b + 3ab2<\/sup> + b3<\/sup>]
\n\u21d2 -x3<\/sup> + 6x2<\/sup> + 14x + 8 = 0
\nis not a Q.E. [\u2235 degree = 3]<\/p>\n

viii) x3<\/sup> – 4x2<\/sup> – x + 1 = (x – 2)3<\/sup>
\nAnswer:
\nGiven : x3<\/sup> – 4x2<\/sup> – x + 1 = (x – 2)3<\/sup>
\n\u21d2 x3<\/sup> – 4x2<\/sup> – x + 1 = x3<\/sup> – 6x2<\/sup> + 12x – 8
\n\u21d2 6x2<\/sup> – 12x + 8 – 4x2<\/sup> – x + 1 = 0
\n\u21d2 2x2<\/sup> – 13x + 9 = 0 is a Q.E.<\/p>\n

Question 2.
\nRepresent the following situations in the form of quadratic equations:
\ni) The area of a rectangular plot is 528 m2<\/sup>. The length of the plot (in meters) is one more than twice its breadth. We need to find the length and breadth of the plot.
\nAnswer:
\nLet the breadth of the rectangular plot be x m.
\nThen its length (by problem) = 2x + 1.
\nArea = l . b = (2x + 1) . x = 2x2<\/sup> + x
\nBut area = 528 m2<\/sup> (\u2235 given)
\n\u2234 2x2<\/sup> + x = 528
\n\u21d2 2x2<\/sup> + x – 528 = 0 where x is the breadth of the rectangle.<\/p>\n

ii) The product of two consecutive positive integers is 306. We need to find the integers.
\nAnswer:
\nLet the consecutive integers be x and x + 1.
\nTheir product = x(x + 1) = x2<\/sup> + x
\nBy problem x2<\/sup> + x = 306
\n\u21d2 x2<\/sup> + x – 306 = 0
\nwhere x is the smaller integer.<\/p>\n

iii) Rohan’s mother is 26 years older than him. The product of their ages after 3 years will be 360 years. We need to find Rohan’s present age.
\nAnswer:
\nLet the present age of Rohan be x years.
\nThen age of Rohan’s mother = x + 26
\nAfter 3 years:
\nAge of Rohan would be = x + 3
\nRohan’s mother’s age would be = (x + 26) + 3 = x + 29
\nBy problem (x + 3) (x + 29) = 360
\n\u21d2 x(x + 29) + 3(x + 29) = 360
\n\u21d2 x2<\/sup> + 29x + 3x + 87 = 360
\n\u21d2 x2<\/sup> + 32x + 87 – 360 = 0
\n\u21d2 x2<\/sup> + 32x – 273 = 0
\n\u21d2 x2<\/sup> + 39x – 7x – 273 = 0
\n\u21d2 x (x + 39) – 7 (x + 39) = 0
\n\u21d2 (x – 7) (x + 39) = 0
\n\u21d2 x = 7 or x = -39 ‘x’ being age cannot be negative.
\n\u2234 x = Present age of Rohan = 7 years.<\/p>\n

\"AP<\/p>\n

iv) A train travels a distance of 480 km at a uniform speed. If the speed had been 8km\/h less, then it would have taken 3 hours more to cover the same distance. We need to find the speed of the train.
\nAnswer:
\nLet the speed of the train be x km\/h.
\nThen time taken to travel a distance of distance of 480 km = \\(\\frac{\\text { distance }}{\\text { speed }}\\) = \\(\\frac{480}{x}\\)
\nIf the speed is 8km\/h less, then time needed to cover the same distance would be \\(\\frac{480}{x-8}\\)
\n\"AP
\n\u21d2 x2<\/sup> – 8x = 1280
\n\u21d2 x2<\/sup> – 8x – 1280 = 0
\nwhere x is the speed of the train.<\/p>\n","protected":false},"excerpt":{"rendered":"

AP State Board Syllabus\u00a0AP SSC 10th Class Maths Textbook Solutions Chapter 5 Quadratic Equations Ex 5.1 Textbook Questions and Answers. AP State Syllabus SSC 10th Class Maths Solutions 5th Lesson Quadratic Equations Exercise 5.1 10th Class Maths 5th Lesson Quadratic Equations Ex 5.1 Textbook Questions and Answers Question 1. Check whether the following are quadratic … Read more<\/a><\/p>\n","protected":false},"author":5,"featured_media":35172,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[2],"tags":[],"jetpack_featured_media_url":"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2023\/02\/Chapter-5-Quadratic-Equations-Ex-5.1.png","_links":{"self":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts\/4365"}],"collection":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/comments?post=4365"}],"version-history":[{"count":1,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts\/4365\/revisions"}],"predecessor-version":[{"id":34167,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts\/4365\/revisions\/34167"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/media\/35172"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/media?parent=4365"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/categories?post=4365"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/tags?post=4365"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}