{"id":1510,"date":"2020-09-15T12:29:21","date_gmt":"2020-09-15T06:59:21","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=1510"},"modified":"2021-07-02T15:01:50","modified_gmt":"2021-07-02T09:31:50","slug":"kseeb-sslc-class-10-maths-solutions-chapter-10-ex-10-4","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-sslc-class-10-maths-solutions-chapter-10-ex-10-4\/","title":{"rendered":"KSEEB SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Ex 10.4"},"content":{"rendered":"

KSEEB SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Ex 10.4<\/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 10.4.<\/p>\n

Karnataka SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Exercise 10.4<\/h2>\n

Question 1.
\nFind the nature of the roots of the following quadratic equations. If the real roots exist, find them :
\n(i) 2x2<\/sup> – 3x + 5 = 0
\n(ii) 3x2<\/sup> – \\(4 \\sqrt{3} x\\) + 4 = 0
\n(iii) 2x2<\/sup> – 6x + 3 = 0
\nSolution:
\ni) 2x2<\/sup> – 3x + 5 = 0
\nIt is in the form of ax2<\/sup> + bx + c = 0
\na = 2, b = – 3 and c = 5
\nNature of roots = b2<\/sup> – 4ac
\nDiscriminant = (- 3)2<\/sup> – 4 \u00d7 (2)(5)
\n= 9 – 40
\n= – 31
\nDiscriminant = – 31 < 0
\nThe given Quadratic equation has no real roots.<\/p>\n

(ii) 3x2<\/sup> – \\(4 \\sqrt{3} x\\) + 4 = 0
\nHere, a = 3, b = \\(4 \\sqrt{3}\\), c = 4
\n\"KSEEB<\/p>\n

(iii) 2x2<\/sup> – 6x + 3 = 0
\nHere, a = 2, b = -6, c = 3
\nb2<\/sup> – 4ac = (-6)2<\/sup> – 4(2)(3)
\n= 36 – 24 = 12
\n\u2234 b2<\/sup> – 4ac = 12 > 0.
\n\u2234 It has two distinct roots.
\n2x2<\/sup> – 6x + 3 = 0
\nHere, a = 2, b = -3, c = 3
\n\"KSEEB<\/p>\n

\"KSEEB<\/p>\n

Question 2.
\nFind the values of k for each of the following quadratic equations, so that they have two equal roots
\n(i) 2x2<\/sup> + kx+3 = 0
\n(ii) kx(x – 2) + 6 = 0
\nSolution:
\n(i) Comparing the given quadratic equation with ax2+ bx + c = 0, we get a = 2,b = k,c = 3
\n\u2234 b2<\/sup> – 4ac = (k)2<\/sup> – 4(2)(3) = k2<\/sup> – 24
\n\u2235 For a quadratic equation to have equal roots, b2<\/sup> – 4ac = 0
\n\"KSEEB<\/p>\n

(ii) kx(x – 2) + 6 = 0 \u21d2 kx2<\/sup> – 2kx + 6 = 0
\nComparing kx2<\/sup> – 2kx + 6 = 0 with ax2<\/sup> + bx + c = 0, we get
\na = k, b = -2k, c = 6
\n\u2234\u00a0 b2<\/sup> – 4ac = (-2k)2<\/sup> – 4(k)(6) = 4k2<\/sup> – 24k
\nSince, the roots are equal
\n\"KSEEB
\nBut k cannot be 0, otherwise, the given equation is not quadratic. Thus, the required value of k is 6.<\/p>\n

\"KSEEB<\/p>\n

Question 3.
\nIs it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m2<\/sup> ? If so, find its length and breadth.
\n\"KSEEB
\nSolution:
\nLet the breadth of rectangular mango grove is ‘x’ and length of rectangular mango grove is 2x
\nArea of rectangular mango grove = 800 m2<\/sup>
\nlength \u00d7 breadth = 800
\nx (2x)= 800
\n2x2<\/sup> = 800
\nx2<\/sup> = \\(\\frac{800}{2}\\) = 400
\nx = \u00b1 \\(\\sqrt{400}\\)
\nx = \u00b1 20m
\n\u2234 breadth of mango grove x = 20 m
\nlength of mango grove = 2x = 2 \u00d7 20 = 40 m
\nIt is possible to construct a rectangular mango grove of length 40m and breadth 20m.<\/p>\n

\"KSEEB<\/p>\n

Question 4.
\nIs the following situation possible? If so, determine their present ages. The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48.
\nSolution:
\nLet the age of one friend = x years
\n\u2234 Age of other friend = (20 – x) years
\nFour years ago,
\nAge of one friend = (x – 4) years
\nAge of other friend = (20 – x – 4) years
\n= (16 -x) years
\nAccording to the condition,
\n(x – 4) \u00d7 (16 – x) = 48
\n\u21d2 16x – 64 – x2<\/sup> + 4x = 48
\n\u21d2 – x2<\/sup> + 20x – 64 – 48 = 0
\n\u21d2 -x2<\/sup> + 20x – 112 = 0
\n\u21d2 x2<\/sup> – 20x + 112 = 0 …(1)
\nComparing equation (1) with ax2<\/sup> + bx + c = 0, we get a = 1, b = – 20 and c = 112
\n\"KSEEB
\nSince, b2<\/sup> – 4ac is less than 0.
\n\u2234 The quadratic equation (1) has no real roots.
\nThus, the given situation is not possible.<\/p>\n

\"KSEEB<\/p>\n

Question 5.
\nIs it possible to design a rectangular park of perimeter 80 m. and area 400 m2<\/sup>? If so, find its length and breadth.
\nSolution:
\n\"KSEEB
\nLet the length of rectangular park is ‘x’
\nPerimeter of the rectangular park = 80m
\n2 (length + breadth) = 80
\nlength + breadth = \\(\\frac{80}{2}\\) = 40
\nbreadth = (40 – x)
\narea of rectangle = l \u00d7 b
\nx (40 – x) = 400
\n40x – x2<\/sup> = 400
\nx2<\/sup> – 40x + 400 = 0
\nx2<\/sup> – 20x – 20x + 400 = 0
\nx (x – 20) – 20 (x – 20) = 0
\n(x – 20) (x – 20) = 0
\nx – 20 (or) x – 20 = 0
\nx = 20 m
\nlength of rectangular park = 20 m and breadth of rectangular park = (40 – x)
\n= 40 – 20 = 20m
\n\u2234 It is possible to design rectangular park of length 20 m and breadth is 20 m.
\n\u2234 The park is a square having 20m side.<\/p>\n

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Ex 10.4 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Exercise 10.4, 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 10.4 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Exercise 10.4. Karnataka SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Exercise 10.4 Question 1. Find the nature of …<\/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\/1510"}],"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=1510"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/1510\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=1510"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=1510"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=1510"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}