{"id":15494,"date":"2024-02-29T05:24:50","date_gmt":"2024-02-28T23:54:50","guid":{"rendered":"https:\/\/www.aplustopper.com\/?p=15494"},"modified":"2024-02-29T14:57:27","modified_gmt":"2024-02-29T09:27:27","slug":"selina-icse-solutions-class-10-maths-quadratic-equations","status":"publish","type":"post","link":"https:\/\/www.aplustopper.com\/selina-icse-solutions-class-10-maths-quadratic-equations\/","title":{"rendered":"Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations"},"content":{"rendered":"
Selina Publishers Concise Mathematics Class 10 ICSE Solutions Chapter 5\u00a0Quadratic Equations<\/strong><\/p>\n Find which of the following equations are quadratic:<\/p>\n Solution 1(i) Solution 1(ii) Solution 1(iii) Solution 1(iv) Solution 1(v) Solution 1(vi) Question 2(i) Question 2(ii). Question 3. Question 4. Question 5. Question 1.<\/strong> Question 2.<\/strong> Question 3.<\/strong> Question 4.<\/strong> Question 5.<\/strong> Question 6.<\/strong> Question 1.<\/strong> Question 2.<\/strong> Question 3.<\/strong> Question 4.<\/strong> Question 5.<\/strong> Question 6.<\/strong> Question 7.<\/strong> Question 8.<\/strong> Question 9.<\/strong> Question 10.<\/strong> Question 11.<\/strong> Question 12.<\/strong> Question 13.<\/strong> Question 14.<\/strong> Question 15.<\/strong> Question 16.<\/strong> Question 17.<\/strong> Question 18.<\/strong> Question 19.<\/strong> Question 20.<\/strong> Question 21.<\/strong> Question 22.<\/strong> Question 23.<\/strong> Question 24.<\/strong> Question 25.<\/strong> Question 26.<\/strong> Question 27.<\/strong> Question 28.<\/strong> Question 29.<\/strong> Question 30.<\/strong> Question 31.<\/strong> Question 32.<\/strong> Question 33.<\/strong> Question 34.<\/strong> Question 35. Question 1.<\/strong> Question 2.<\/strong> Question 3(i).<\/strong> Question 3(ii).<\/strong> Question 3(iii).<\/strong> Question 3(iv).<\/strong> Question 4.<\/strong> Question 5.<\/strong> Question 6.<\/strong> Question 7.<\/strong>Quadratic Equations Exercise 5A – Selina Concise Mathematics Class 10 ICSE Solutions<\/h3>\n
\n(3x – 1)2<\/sup>\u00a0= 5(x + 8)
\n\u21d2\u00a0(9x2<\/sup>\u00a0– 6x + 1) = 5x + 40
\n\u21d2\u00a09x2<\/sup>\u00a0– 11x – 39 =0; which is of the form ax2<\/sup>\u00a0+ bx + c = 0.
\n\u2234\u00a0Given equation is a quadratic equation.<\/p>\n
\n5x2<\/sup>\u00a0– 8x = -3(7 – 2x)
\n\u21d2\u00a05x2<\/sup>\u00a0– 8x = 6x – 21
\n\u21d2\u00a05x2<\/sup>\u00a0– 14x + 21 =0; which is of the form ax2<\/sup>\u00a0+ bx + c = 0.
\n\u2234\u00a0Given equation is a quadratic equation.<\/p>\n
\n(x – 4)(3x + 1) = (3x – 1)(x +2)
\n\u21d2\u00a03x2<\/sup>\u00a0+ x – 12x – 4 = 3x2<\/sup>\u00a0+ 6x – x – 2
\n\u21d2\u00a016x + 2 =0; which is not of the form ax2<\/sup>\u00a0+ bx + c = 0.
\n\u2234\u00a0Given equation is not a quadratic equation.<\/p>\n
\nx2<\/sup>\u00a0+ 5x – 5 = (x – 3)2
\n<\/sup>\u21d2\u00a0x2<\/sup>\u00a0+ 5x – 5 = x2<\/sup>\u00a0– 6x + 9
\n\u21d2\u00a011x – 14 =0; which is not of the form ax2<\/sup>\u00a0+ bx + c = 0.
\n\u2234\u00a0Given equation is not a quadratic equation.<\/p>\n
\n7x3<\/sup>\u00a0– 2x2<\/sup>\u00a0+ 10 = (2x – 5)2
\n<\/sup>\u21d2\u00a07x3<\/sup>\u00a0– 2x2<\/sup>\u00a0+ 10 = 4x2<\/sup>\u00a0– 20x + 25
\n\u21d2\u00a07x3<\/sup>\u00a0– 6x2<\/sup>\u00a0+ 20x – 15 = 0; which is not of the form ax2<\/sup>\u00a0+ bx + c = 0.
\n\u2234\u00a0Given equation is not a quadratic equation.<\/p>\n
\n(x – 1)2<\/sup>\u00a0+ (x + 2)2<\/sup>\u00a0+ 3(x +1) = 0
\n\u21d2\u00a0x2<\/sup>\u00a0– 2x + 1 + x2<\/sup>\u00a0+ 4x + 4 + 3x + 3 = 0
\n\u21d2\u00a02x2<\/sup>\u00a0+ 5x + 8 = 0; which is of the form ax2<\/sup>\u00a0+ bx + c = 0.
\n\u2234\u00a0Given equation is a quadratic equation.<\/p>\n
\nIs x = 5 a solution of the quadratic equation x2<\/sup>\u00a0– 2x – 15 = 0?
\nSolution:
\nx2<\/sup>\u00a0– 2x – 15 = 0
\nFor x = 5 to be solution of the given quadratic equation it should satisfy the equation.
\nSo, substituting x = 5 in the given equation, we get
\nL.H.S = (5)2<\/sup>\u00a0– 2(5) – 15
\n= 25 – 10 – 15
\n= 0
\n= R.H.S
\nHence, x = 5 is a solution of the quadratic equation x2<\/sup>\u00a0– 2x – 15 = 0.<\/p>\n
\nIs x = -3 a solution of the quadratic equation 2x2<\/sup>\u00a0– 7x + 9 = 0?
\nSolution:
\n2x2<\/sup>\u00a0– 7x + 9 = 0
\nFor x = -3 to be solution of the given quadratic equation it should satisfy the equation
\nSo, substituting x = 5 in the given equation, we get
\nL.H.S =2(-3)2<\/sup>\u00a0– 7(-3) + 9
\n= 18 + 21 + 9
\n=\u00a048
\n\u2260\u00a0R.H.S
\nHence, x = -3 is not a solution of the quadratic equation 2x2<\/sup>\u00a0– 7x + 9 = 0.<\/p>\n
\nIf\u00a0\\(\\sqrt{\\frac{2}{3}}\\)\u00a0is a solution of equation 3x2<\/sup>\u00a0+ mx + 2 = 0, find the value of m.
\nSolution:
\nFor x = \\(\\sqrt{\\frac{2}{3}}\\) to be solution of the given quadratic equation it should satisfy the equation
\nSo, substituting x = \\(\\sqrt{\\frac{2}{3}}\\) in the given equation, we get
\n<\/p>\n
\n\\(\\frac{2}{3}\\)\u00a0and\u00a01 are the solutions of equation mx2<\/sup>\u00a0+\u00a0nx\u00a0+ 6 = 0. Find the values of m and n.
\nSolution:
\nFor x =\u00a0\u00a0\\(\\frac{2}{3}\\) and x = 1 to be solutions of the given quadratic equation it should satisfy the equation
\nSo, substituting x =\u00a0\u00a0\\(\\frac{2}{3}\\) and x = 1 in the given equation, we get
\n
\nSolving equations (1) and (2) simultaneously,
\n4m\u00a0 + 6n + 54 = 0 …..(1)
\nm + n\u00a0 + 6 = 0 ….(2)
\n(1) – (2) \u00d7 6
\n\u21d2 -2m + 18 = 0
\n\u21d2 m = 9
\nSubstitute in (2)
\n\u21d2 n = -15<\/p>\n
\nIf 3 and -3 are the solutions of equation ax2<\/sup>\u00a0+ bx – 9 = 0. Find the values of a and b.
\nSolution:
\nFor x = 3 and x = -3 to be solutions of the given quadratic equation it should satisfy the equation
\nSo, substituting x = 3 and x = -3 in the given equation, we get
\n
\nSolving equations (1) and (2) simultaneously,
\n9a + 3b – 9 = 0 …(1)
\n9a – 3b – 9 = 0 …(2)
\n(1) + (2)
\n\u21d2 18a – 18 = 0
\n\u21d2 a = 1
\nSubstitute in (2)
\n\u21d2 b = 0<\/p>\nQuadratic Equations Exercise 5B – Selina Concise Mathematics Class 10 ICSE Solutions<\/h3>\n
\nWithout solving, comment upon the nature of roots of each of the following equations :
\n(i) 7x2<\/sup>\u00a0– 9x +2 =0
\n(ii) 6x2<\/sup>\u00a0– 13x +4 =0
\n(iii) 25x2<\/sup>\u00a0– 10x +1=0
\n(iv)\u00a0x2<\/sup>\u00a0+ 2\u221a3x – 9=0
\n(v) x2<\/sup>\u00a0– ax – b2<\/sup>\u00a0=0
\n(vi) 2x2<\/sup>\u00a0+8x +9=0
\nSolution:<\/strong>
\n<\/p>\n
\nFind the value of p, if the following quadratic equation has equal roots : 4x2<\/sup>\u00a0– (p – 2)x + 1 = 0
\nSolution:<\/strong>
\n<\/p>\n
\nFind the value of ‘p’, if the following quadratic equations have equal roots : x2<\/sup>\u00a0+ (p – 3)x + p = 0
\nSolution:<\/strong>
\nx2<\/sup>\u00a0+ (p – 3)x + p = 0
\nHere, a = 1, b = (p – 3), c = p
\nSince, the roots are equal,
\n\u21d2\u00a0b2<\/sup>– 4ac = 0
\n\u21d2\u00a0(p – 3)2<\/sup>– 4(1)(p) = 0
\n\u21d2p2<\/sup>\u00a0+ 9 – 6p – 4p = 0
\n\u21d2\u00a0p2<\/sup>– 10p + 9 = 0
\n\u21d2p2<\/sup>-9p – p + 9 = 0
\n\u21d2p(p – 9) – 1(p – 9) = 0
\n\u21d2\u00a0(p -9)(p – 1) = 0
\n\u21d2\u00a0p – 9 = 0 or p – 1 = 0
\n\u21d2\u00a0p = 9 or p = 1<\/p>\n
\nThe equation 3x2<\/sup>\u00a0– 12x + (n – 5)=0 has equal roots. Find the value of n.
\nSolution:<\/strong>
\n<\/p>\n
\nFind the value of m, if the following equation has equal roots : (m – 2)x2<\/sup>\u00a0– (5+m)x +16 =0
\nSolution:<\/strong>
\n<\/p>\n
\nFind the value of p for which the equation 3x2<\/sup>– 6x + k = 0 has distinct and real roots.
\nSolution:<\/strong>
\n<\/p>\nQuadratic Equations Exercise 5C – Selina Concise Mathematics Class 10 ICSE Solutions<\/h3>\n
\nSolve : x\u00b2 – 10x – 24 = 0
\nSolution:<\/strong>
\n<\/p>\n
\nSolve : x\u00b2 – 16 = 0
\nSolution:<\/strong>
\n<\/p>\n
\n
\nSolution:<\/strong>
\n<\/p>\n
\nSolve : x(x – 5) = 24
\nSolution:<\/strong>
\n<\/p>\n
\n
\nSolution:<\/strong>
\n<\/p>\n
\n
\nSolution:<\/strong>
\n<\/p>\n
\n
\nSolution:<\/strong>
\n<\/p>\n
\n
\nSolution:<\/strong>
\n<\/p>\n
\nSolve : (2x – 3)\u00b2 = 49
\nSolution:<\/strong>
\n<\/p>\n
\nSolve : 2(x\u00b2 – 6) = 3(x – 4)
\nSolution:<\/strong>
\n<\/p>\n
\nSolve : (x + 1)(2x + 8) = (x + 7)(x + 3)
\nSolution:<\/strong>
\n<\/p>\n
\nSolve : x\u00b2 – (a + b)x + ab = 0
\nSolution:<\/strong>
\n<\/p>\n
\n(x + 3)\u00b2 – 4(x + 3) – 5 = 0
\nSolution:<\/strong>
\n<\/p>\n
\n4(2x – 3)\u00b2 – (2x – 3) – 14 = 0
\nSolution:<\/strong>
\n<\/p>\n
\n
\nSolution:<\/strong>
\n<\/p>\n
\n2x2<\/sup>\u00a0– 9x + 10 = 0, When
\n(i)\u00a0x\u2208\u00a0N
\n(ii)\u00a0x\u2208\u00a0Q
\nSolution:<\/strong>
\n<\/p>\n
\n
\nSolution:<\/strong>
\n<\/p>\n
\n
\nSolution:<\/strong>
\n<\/p>\n
\n
\nSolution:<\/strong>
\n<\/p>\n
\n
\nSolution:<\/strong>
\n<\/p>\n
\nFind the quadratic equation, whose solution set is :
\n(i) {3, 5} (ii) {-2, 3}
\nSolution:<\/strong>
\n<\/p>\n
\n
\nSolution:<\/strong>
\n<\/p>\n
\n
\nSolution:<\/strong>
\n<\/p>\n
\nFind the value of x, if a + 1=0 and x2<\/sup>\u00a0+ ax – 6 =0.
\nSolution:<\/strong>
\n<\/p>\n
\nFind the value of x, if a + 7=0; b + 10=0 and 12x2<\/sup>\u00a0= ax – b.
\nSolution:<\/strong>
\nIf a + 7 =0, then a = -7
\nand b + 10 =0, then b = – 10
\nPut these values of a and b in the given equation
\n<\/p>\n
\nUse the substitution y= 2x +3 to solve for x, if 4(2x+3)2<\/sup>\u00a0– (2x+3) – 14 =0.
\nSolution:<\/strong>
\n<\/p>\n
\nWithout solving the quadratic equation 6x2<\/sup>\u00a0– x – 2=0, find whether x = 2\/3\u00a0is a solution of this equation or not.
\nSolution:<\/strong>
\n<\/p>\n
\nDetermine whether x = -1 is a root of the equation x2<\/sup>\u00a0– 3x +2=0 or not.
\nSolution:<\/strong>
\nx2<\/sup> – 3x +2=0
\nPut x = -1 in L.H.S.
\nL.H.S. = (-1)2\u00a0<\/sup>– 3(-1) +2
\n= 1 +3 +2=6 \u2260 R.H.S
\nThen x = -1 is not the solution of the given equation.<\/p>\n
\nIf x = 2\/3 is a solution of the quadratic equation 7x2<\/sup>+mx – 3=0; Find the value of m.
\nSolution:<\/strong>
\n<\/p>\n
\nIf x = -3 and x = 2\/3\u00a0are solutions of quadratic equation mx2\u00a0<\/sup>+ 7x + n = 0, find the values of m and n.
\nSolution:<\/strong>
\n<\/p>\n
\nIf quadratic equation x2<\/sup>\u00a0– (m + 1) x + 6=0 has one root as x =3; find the value of m and the root of the equation.
\nSolution:<\/strong>
\n<\/p>\n
\nGiven that 2 is a root of the equation 3x\u00b2 – p(x + 1) = 0 and that the equation px\u00b2 – qx + 9 = 0 has equal roots, find the values of p and q.
\nSolution:<\/strong>
\n<\/p>\n
\n
\nSolution:<\/strong>
\n<\/p>\n
\n
\nSolution:<\/strong>
\n<\/p>\n
\n<\/strong>If -1 and 3 are the roots of x2<\/sup> + px + q = 0, find the values of p and q.
\nSolution:<\/strong>
\n<\/p>\nQuadratic Equations Exercise 5D – Selina Concise Mathematics Class 10 ICSE Solutions<\/h3>\n
\n
\nSolution:<\/strong>
\n
\n
\n
\n
\n
\n
\n<\/p>\n
\nSolve each of the following equations for x and give, in each case, your answer correct to one decimal place :
\n(i) x2<\/sup>\u00a0– 8x+5=0
\n(ii) 5x2<\/sup>\u00a0+10x – 3 =0
\nSolution:<\/strong>
\n<\/p>\n
\nSolve each of the following equations for x and give, in each case, your answer correct to two decimal places :
\n(i) 2x2<\/sup>\u00a0– 10x +5=0
\nSolution:<\/strong>
\n<\/p>\n
\nSolve each of the following equations for x and give, in each case, your answer correct to two decimal places :
\n4x + 6\/x + 13 = 0
\nSolution:<\/strong>
\n<\/p>\n
\nSolve each of the following equations for x and give, in each case, your answer correct to two decimal places :
\nx2<\/sup>\u00a0– 3x – 9 =0
\nSolution:<\/strong>
\n<\/p>\n
\nSolve each of the following equations for x and give, in each case, your answer correct to two decimal places :
\nx2<\/sup>\u00a0– 5x – 10 = 0
\nSolution:<\/strong>
\n<\/p>\n
\nSolve each of the following equations for x and give, in each case, your answer correct to 3 decimal places :
\n(i) 3x2<\/sup>\u00a0– 12x – 1 =0
\n(ii) x2<\/sup>\u00a0– 16 x +6= 0
\n(iii) 2x2<\/sup>\u00a0+ 11x + 4= 0
\nSolution:<\/strong>
\n<\/p>\n
\nSolve:
\n(i) x4<\/sup>\u00a0– 2x2<\/sup>\u00a0– 3 =0
\n(ii) x4<\/sup>\u00a0– 10x2<\/sup>\u00a0+9 =0
\nSolution:<\/strong>
\n<\/p>\n
\nSolve :
\n(i) (x2<\/sup>\u00a0– x)2<\/sup>\u00a0+ 5(x2<\/sup>\u00a0– x)+ 4=0
\n(ii) (x2<\/sup>\u00a0– 3x)2<\/sup>\u00a0– 16(x2<\/sup>\u00a0– 3x) – 36 =0
\nSolution:<\/strong>
\n<\/p>\n