{"id":9970,"date":"2024-02-24T09:45:21","date_gmt":"2024-02-24T04:15:21","guid":{"rendered":"https:\/\/tsboardsolutions.in\/?p=9970"},"modified":"2024-02-26T09:44:43","modified_gmt":"2024-02-26T04:14:43","slug":"ts-10th-class-maths-solutions-chapter-5-ex-5-1","status":"publish","type":"post","link":"https:\/\/tsboardsolutions.in\/ts-10th-class-maths-solutions-chapter-5-ex-5-1\/","title":{"rendered":"TS 10th Class Maths Solutions Chapter 5 Quadratic Equations Ex 5.1"},"content":{"rendered":"

Students can practice 10th Class Maths Solutions Telangana<\/a> Chapter 5 Quadratic Equations Ex 5.1 to get the best methods of solving problems.<\/p>\n

TS 10th Class Maths Solutions Chapter 5 Quadratic Equations Exercise 5.1<\/h2>\n

Question 1.
\nCheck whether the following are quadratic equations or not.
\ni) (x + 1)2<\/sup> = 2(x – 3)
\nSolution:
\nGiven: (x + 1)2<\/sup> = 2(x – 3)
\n\u21d2 x2<\/sup> + 2x + 1 = 2(x – 3)
\n\u21d2 x2<\/sup> + 2x + 1 – 2x + 6 = 0
\n\u21d2 x2<\/sup> + 7 = 0 is a Q.E.<\/p>\n

ii) x2<\/sup> – 2x = -2(3 – x)
\nSolution:
\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

iii) (x – 2) (x + 1) = (x – 1) (x + 3)
\nSolution:
\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<\/sup> 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

\"TS<\/p>\n

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

v) (2x – 1)(x – 3) = (x + 5)(x – 1)
\nSolution:
\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
\nHence it is a Q.E.<\/p>\n

vi) x2<\/sup> + 3x + 1 = (x – 2)2<\/sup>
\nSolution:
\nGiven : x2<\/sup> + 3x + 1 = (x – 2)2<\/sup>
\n\u21d2 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

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

viii) x3<\/sup> – 4x2<\/sup> – x + 1 = (x – 2)3<\/sup>
\nSolution:
\nGiven : x3<\/sup> – 4x2<\/sup> – x + 1 = (x – 2)3<\/sup>
\n\u21d2 x3<\/sup> – 4x2<\/sup> – x + 1 = (x – 2)3<\/sup>
\n= 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 metres) is one more than twice its breadth. We need to find the length and breadth of the plot.
\nSolution:
\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 but 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 the consecutive positive integers is 306. We need to find the integers.
\nSolution:
\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

\"TS<\/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.
\nSolution:
\nLet the present age of Rohan be x years. Then 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
\nWhere x is Rohan’s present age.<\/p>\n

iv) A train travels a distance of 480 km at a uniform speed. If the speed had been 8km\/hr less, then it would have taken 3 hours more to cover the same distance. We need to find the speed of the train.
\nSolution:
\nLet the speed of the train be x kmph. Then time taken to travel a distance
\n\"TS
\nIt the speed is 8km\/hr less, then time needed to cover the same distance
\nwould be \\(\\frac{480}{x-8}\\)
\nBy problem = \\(\\frac{480}{x-8}\\) – \\(\\frac{480}{x}\\) = 3
\n\"TS
\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":"

Students can practice 10th Class Maths Solutions Telangana Chapter 5 Quadratic Equations Ex 5.1 to get the best methods of solving problems. TS 10th Class Maths Solutions Chapter 5 Quadratic Equations Exercise 5.1 Question 1. Check whether the following are quadratic equations or not. i) (x + 1)2 = 2(x – 3) Solution: Given: (x … Read more<\/a><\/p>\n","protected":false},"author":3,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[16],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/9970"}],"collection":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/users\/3"}],"replies":[{"embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/comments?post=9970"}],"version-history":[{"count":5,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/9970\/revisions"}],"predecessor-version":[{"id":10084,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/9970\/revisions\/10084"}],"wp:attachment":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/media?parent=9970"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/categories?post=9970"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/tags?post=9970"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}