{"id":1347,"date":"2020-09-15T10:14:58","date_gmt":"2020-09-15T04:44:58","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=1347"},"modified":"2021-07-02T15:01:50","modified_gmt":"2021-07-02T09:31:50","slug":"kseeb-sslc-class-10-maths-solutions-chapter-10-ex-10-1","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-sslc-class-10-maths-solutions-chapter-10-ex-10-1\/","title":{"rendered":"KSEEB SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Ex 10.1"},"content":{"rendered":"

KSEEB SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Ex 10.1<\/strong> are part of KSEEB SSLC Class 10 Maths Solutions<\/a>. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Exercise 10.1.<\/p>\n

Karnataka SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Exercise 10.1<\/h2>\n

Question 1.
\nCheck whether the following are quadratic equations:
\ni) (x + 1)2<\/sup> = 2 (x – 3)
\nii) x2<\/sup> – 2x = (-2) (3 – x)
\niii) (x – 2) (x + 1) = (x – 1) (x + 3)
\niv) (x – 3) (2x + 1) = x (x + 5)
\nv) (2x – 1) (x – 3) = (x +5) (x – 1)
\nvi) x2<\/sup> + 3x + 1 = (x – 2)2<\/sup>
\nvii) (x + 2)3<\/sup> = 2x (x2<\/sup> – 1)
\nviii) x3<\/sup> – 4x2<\/sup> – x + 1 = (x – 2)3<\/sup>
\nSolution:
\ni) (x + 1)2<\/sup> = 2(x – 3)
\nx2<\/sup> + 2x + 1 = 2x – 6
\nx2<\/sup> + 2x – 2x + 1 – 6 = 0
\nx2<\/sup> – 5 = 0
\nIt is in the form of ax2<\/sup> + c = 0
\nTherefore, the given equation is a Quadratic equation.<\/p>\n

ii) x2<\/sup> – 2x = (- 2) (3 – x)
\nx2<\/sup> – 2x = – 6 + 2x
\nx2<\/sup> – 2x – 2x + 6 = 0
\nx2<\/sup> – 4x + 6 = 0
\nIt is in the form of ax2 + bx + c = 0
\nTherefore, the given equation is a Quadratic equation.<\/p>\n

iii) (x – 2) (x + 1) = (x – 1) (x + 3)
\nx2<\/sup> + x – 2x – 2 = x2<\/sup> + 3x – x – 3
\nx2<\/sup> – x – 2 = x2<\/sup> + 2x – 3
\nx2<\/sup> – x2<\/sup> – x – 2x – 2 + 3 = 0
\n– 3x + 1 = 0
\nIt is in not in the form of ax2<\/sup> + bx + c = 0
\nTherefore, the given equation is not a Quadratic equation.<\/p>\n

iv) (x – 3) (2x + 1) = x (x + 5)
\n2x2<\/sup> + x – 6x – 3 = x2 + 5x
\n2x2<\/sup> – 5x – 3 = x2<\/sup> + 5x
\n2x2<\/sup> – x2<\/sup> – 5x – 5x – 3 = 0
\nx2<\/sup> – 10x – 3 = 0
\nIt is in the form of ax2<\/sup> + bx + c = 0
\nTherefore, the- given equation is a Quadratic equation.<\/p>\n

v) (2x – 1) (x – 3) = (x + 5) (x – 1)
\n2x2<\/sup> – 6x – x + 3 = x2<\/sup> – x + 5x – 5
\n2x2<\/sup> – 7x + 3 = x2<\/sup> + 4x – 5
\n2x2<\/sup> – x2<\/sup> – 7x – 4x + 3 + 5 = 0
\nx2<\/sup> – 11x + 8 = 0
\nIt is in the form of ax2<\/sup> + bx + c = 0
\nTherefore, the given equation is a Quadratic equation.<\/p>\n

vi) x2<\/sup> + 3x + 1 = (x – 2)2<\/sup>
\nx2<\/sup> + 3x + 1 = x2<\/sup> – 4x + 4
\nx2<\/sup> – x2<\/sup> + 3x + 4x + 1 – 4 = 0
\n7x – 3 = 0
\nIt is not in the form of ax2<\/sup> + bx + c = 0
\nTherefore, the given equation is not a Quadratic equation.<\/p>\n

vii) (x + 2)3<\/sup> = 2x (x2<\/sup> – 1)
\nx3<\/sup> + (3)3<\/sup> + 3 \u00d7 2 \u00d7 x (x + 2) = 2x3<\/sup> – 2x
\nx3<\/sup> + 27 + 6x (x + 2) = 2x3<\/sup> – 2x
\nx3<\/sup> + 27 + 6x2<\/sup> + 12x = 2x3<\/sup> – 2x
\n2x3<\/sup> – x3<\/sup> – 6x2<\/sup> – 2x – 12x – 27 = 0
\nx3<\/sup> – 6x2<\/sup> – 14x – 27 = 0
\nIt is not in the form of ax2<\/sup> + bx + c = 0
\nTherefore, the given equation is not a Quadratic equation.<\/p>\n

viii) x3<\/sup> – 4x2<\/sup> – x + 1 = (x – 2)3<\/sup>
\nx3<\/sup> – 4x2<\/sup> – x + 1 = x3<\/sup> – (2)3<\/sup> – 3 \u00d7 2 \u00d7 x(x – 2)
\nx3<\/sup> – 4x3<\/sup> – x + 1 = x3<\/sup> – 8 – 6x (x – 2)
\nx3<\/sup> – 4x2<\/sup> – x + 1 = x3<\/sup> – 8 – 6x2<\/sup> + 12x
\nx3<\/sup> – x3<\/sup> – 4x2<\/sup> + 6x2<\/sup> – x – 12x + 1 + 8 = 0
\n2x2<\/sup> – 13x + 9 = 0
\nIt is in the form of ax2<\/sup> + bx + c = 0
\nTherefore, the given equation is a Quadratic equation.<\/p>\n

Question 2.
\nReprsent the following situations in the form of quadratic equations :
\n(i) 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 rectangular plot (b) be \u2019x’ m.
\nThen the length of th plot is one more than twice its breadth,
\n\u2234 Length (l)= 2x + 1 m.
\nBut Length \u00d7 Breadth = Area of rectangle
\nl \u00d7 b = A
\n\u2234 x \u00d7 (2x + 1) = 528 sq.m.
\n2x2<\/sup> + x = 528
\n\u2234 2x2<\/sup> + x – 528 = 0 is the required equation.
\nNow, we have to find out the value of ‘x’ :
\n2x2<\/sup> + x – 528 = 0
\n2x2<\/sup> – 32x + 33x – 528 = 0
\n2x(x – 16) + 33(x – 16) = 0
\n(x – 16) (2x + 33) = 0
\nIf x – 16 = 0, then x = 16
\nIf 2x + 33 = 0, then x = -33\/2
\n\u2234 Breadth (b) = 16 m.
\nLength (l) = (2x + 1) = 2(16) + 1 = 32 + 1 = 33m
\n\u2234 Length (l) = 33 m
\nBreadth (b) = 16 m.<\/p>\n

(ii) The product of two consecutive positive integers is 306. We need to find the integers.
\nSolution:
\nLet one positive integer be x’.
\nThe Next integer is (x + 1)
\nTheir product is 306.
\n\u2234 x (x + 1) = 306
\nx2<\/sup> + x = 306
\n\u2234 x2<\/sup> + x – 306= 0. This is required equation.
\nNow, we have to solve for positive integer.
\nx2<\/sup> + x – 306 = 0
\nx2<\/sup> + 18x – 17x – 306 = 0
\nx(x + 18) – 17(x + 18) = 0
\n(x + 18) (x – 17) = 0
\nIf x + 18 = 0, then x = -18
\nIf x – 17 = 0, then x = 17
\n\u2234 x = 18, OR x = 17.<\/p>\n

(iii) Rohan\u2019s mother is 26 years older than him. The product of their ages (in years) 3 years from now will be 360. We would like to find Rohan\u2019s present age.
\nSolution:
\nLet the present age of Rohan be ‘x’, then His mother’s age will be (x + 26)
\nAfter 3 years, Age of Rohan is (x + 3). After 3 years his mother’s age will be
\n= (x + 26 + 3)
\n= (x + 29)
\nThen product of their ages is 360.
\n\u2234 (x + 3) (x + 29) = 360
\nx2<\/sup> + 29x + 3x + 87 = 360
\nx2<\/sup> + 32x + 87 = 360
\nx2<\/sup> + 32x + 87 – 360 = 0
\nx2<\/sup> + 32x – 273 = 0.
\nThis is the required equation.
\nNow, we have to solve for the value of ‘x’:
\nx2<\/sup> + 32x – 273 = 0
\nx2<\/sup> + 39x – 7x – 273 = 0
\nx(x + 39) – 7(x + 39) = 0
\n(x + 39) (x – 6) = 0
\nIf x + 39 = 0, then x = -39
\nIf x – 6 = 0, then x = 7
\nPresent age of Rohan\u2019s mother
\n= x + 26
\n= 7 + 26
\n= 33 years.<\/p>\n

(iv) A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km\/h 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 initial speed of a train be ‘x’ km\/h.
\nTime required to travel x km is 1 hour. Time required to travel 480 km ………?
\n\\(\\frac{480}{x}\\) hr
\nIf its speed decreases to 8 km\/h, then it is (x – 8) km\/h.
\nTime required to cover (x – 8) km is 1 Hr.
\nTime required to cover 480 km ………..?
\n\"KSEEB
\n\u2234 x(3x + 456) = 480 (x – 8)
\n3x2<\/sup> + 456x = 480x + 3840
\n3x2<\/sup> + 456x – 480 x + 3840 = 0
\n3x2<\/sup> – 24x + 3840 = 0
\n\u2234 x2<\/sup> – 8x + 1280 = 0
\nThis is the required equation.
\nNow, we have to solve for x :
\nx2<\/sup> – 8x + 1280 = 0
\nx2<\/sup> – 40x + 32x + 1280 = 0
\nx(x – 40) + 32(x + 40) = 0
\n(x – 40) (x + 32) = 0
\nIf x – 40 = 0, then x = 40
\nIf x + 32 = 0, then x = -32
\n\u2234 Average speed of train is 40 km\/hr.<\/p>\n

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Ex 10.1 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Exercise 10.1, drop a comment below and we will get back to you at the earliest.<\/p>\n","protected":false},"excerpt":{"rendered":"

KSEEB SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Ex 10.1 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Exercise 10.1. Karnataka SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Exercise 10.1 Question 1. Check whether the following …<\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[3],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/1347"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=1347"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/1347\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=1347"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=1347"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=1347"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}