{"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 Concise Mathematics Class 10 ICSE Solutions Quadratic Equations<\/h2>\n

Selina Publishers Concise Mathematics Class 10 ICSE Solutions Chapter 5\u00a0Quadratic Equations<\/strong><\/p>\n

Quadratic Equations Exercise 5A – Selina Concise Mathematics Class 10 ICSE Solutions<\/h3>\n

Find which of the following equations are quadratic:<\/p>\n

Solution 1(i)
\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

Solution 1(ii)
\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

Solution 1(iii)
\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

Solution 1(iv)
\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

Solution 1(v)
\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

Solution 1(vi)
\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

Question 2(i)
\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

Question 2(ii).
\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

Question 3.
\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\"Selina<\/p>\n

Question 4.
\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\"Selina
\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

Question 5.
\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\"Selina
\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>\n

Quadratic Equations Exercise 5B – Selina Concise Mathematics Class 10 ICSE Solutions<\/h3>\n

Question 1.<\/strong>
\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\"Selina\"Selina<\/p>\n

Question 2.<\/strong>
\nFind the value of p, if the following quadratic equation has equal roots : 4x2<\/sup>\u00a0– (p – 2)x + 1 = 0
\nSolution:<\/strong>
\n\"Selina<\/p>\n

Question 3.<\/strong>
\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

Question 4.<\/strong>
\nThe equation 3x2<\/sup>\u00a0– 12x + (n – 5)=0 has equal roots. Find the value of n.
\nSolution:<\/strong>
\n\"Selina<\/p>\n

Question 5.<\/strong>
\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\"Selina<\/p>\n

Question 6.<\/strong>
\nFind the value of p for which the equation 3x2<\/sup>– 6x + k = 0 has distinct and real roots.
\nSolution:<\/strong>
\n\"Selina<\/p>\n

Quadratic Equations Exercise 5C – Selina Concise Mathematics Class 10 ICSE Solutions<\/h3>\n

Question 1.<\/strong>
\nSolve : x\u00b2 – 10x – 24 = 0
\nSolution:<\/strong>
\n\"Selina<\/p>\n

Question 2.<\/strong>
\nSolve : x\u00b2 – 16 = 0
\nSolution:<\/strong>
\n\"Selina<\/p>\n

Question 3.<\/strong>
\n\"Selina
\nSolution:<\/strong>
\n\"\"<\/p>\n

Question 4.<\/strong>
\nSolve : x(x – 5) = 24
\nSolution:<\/strong>
\n\"Selina<\/p>\n

Question 5.<\/strong>
\n\"Selina
\nSolution:<\/strong>
\n\"Selina<\/p>\n

Question 6.<\/strong>
\n\"Selina
\nSolution:<\/strong>
\n\"Selina<\/p>\n

Question 7.<\/strong>
\n\"Selina
\nSolution:<\/strong>
\n\"Selina<\/p>\n

Question 8.<\/strong>
\n\"Selina
\nSolution:<\/strong>
\n\"Selina<\/p>\n

Question 9.<\/strong>
\nSolve : (2x – 3)\u00b2 = 49
\nSolution:<\/strong>
\n\"Selina<\/p>\n

Question 10.<\/strong>
\nSolve : 2(x\u00b2 – 6) = 3(x – 4)
\nSolution:<\/strong>
\n\"Selina<\/p>\n

Question 11.<\/strong>
\nSolve : (x + 1)(2x + 8) = (x + 7)(x + 3)
\nSolution:<\/strong>
\n\"Selina<\/p>\n

Question 12.<\/strong>
\nSolve : x\u00b2 – (a + b)x + ab = 0
\nSolution:<\/strong>
\n\"Selina<\/p>\n

Question 13.<\/strong>
\n(x + 3)\u00b2 – 4(x + 3) – 5 = 0
\nSolution:<\/strong>
\n\"Selina<\/p>\n

Question 14.<\/strong>
\n4(2x – 3)\u00b2 – (2x – 3) – 14 = 0
\nSolution:<\/strong>
\n\"Selina<\/p>\n

Question 15.<\/strong>
\n\"Selina
\nSolution:<\/strong>
\n\"Selina<\/p>\n

Question 16.<\/strong>
\n2x2<\/sup>\u00a0– 9x + 10 = 0, When
\n(i)\u00a0x\u2208\u00a0N
\n(ii)\u00a0x\u2208\u00a0Q
\nSolution:<\/strong>
\n\"Selina<\/p>\n

Question 17.<\/strong>
\n\"Selina
\nSolution:<\/strong>
\n\"Selina<\/p>\n

Question 18.<\/strong>
\n\"Selina
\nSolution:<\/strong>
\n\"Selina<\/p>\n

Question 19.<\/strong>
\n\"Selina
\nSolution:<\/strong>
\n\"Selina<\/p>\n

Question 20.<\/strong>
\n\"Selina
\nSolution:<\/strong>
\n\"Selina<\/p>\n

Question 21.<\/strong>
\nFind the quadratic equation, whose solution set is :
\n(i) {3, 5} (ii) {-2, 3}
\nSolution:<\/strong>
\n\"Selina<\/p>\n

Question 22.<\/strong>
\n\"Selina
\nSolution:<\/strong>
\n\"Selina<\/p>\n

Question 23.<\/strong>
\n\"Selina
\nSolution:<\/strong>
\n\"Selina<\/p>\n

Question 24.<\/strong>
\nFind the value of x, if a + 1=0 and x2<\/sup>\u00a0+ ax – 6 =0.
\nSolution:<\/strong>
\n\"Selina<\/p>\n

Question 25.<\/strong>
\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\"Selina<\/p>\n

Question 26.<\/strong>
\nUse the substitution y= 2x +3 to solve for x, if 4(2x+3)2<\/sup>\u00a0– (2x+3) – 14 =0.
\nSolution:<\/strong>
\n\"Selina<\/p>\n

Question 27.<\/strong>
\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\"Selina<\/p>\n

Question 28.<\/strong>
\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

Question 29.<\/strong>
\nIf x = 2\/3 is a solution of the quadratic equation 7x2<\/sup>+mx – 3=0; Find the value of m.
\nSolution:<\/strong>
\n\"Selina<\/p>\n

Question 30.<\/strong>
\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\"Selina<\/p>\n

Question 31.<\/strong>
\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\"Selina<\/p>\n

Question 32.<\/strong>
\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\"Selina<\/p>\n

Question 33.<\/strong>
\n\"Selina
\nSolution:<\/strong>
\n\"Selina<\/p>\n

Question 34.<\/strong>
\n\"Selina
\nSolution:<\/strong>
\n\"Selina<\/p>\n

Question 35.
\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\"Selina<\/p>\n

Quadratic Equations Exercise 5D – Selina Concise Mathematics Class 10 ICSE Solutions<\/h3>\n

Question 1.<\/strong>
\n\"Selina
\nSolution:<\/strong>
\n\"Selina
\n\"Selina
\n\"Selina
\n\"Selina
\n\"Selina
\n\"Selina
\n\"Selina<\/p>\n

Question 2.<\/strong>
\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\"Selina<\/p>\n

Question 3(i).<\/strong>
\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\"Selina<\/p>\n

Question 3(ii).<\/strong>
\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\"Selina<\/p>\n

Question 3(iii).<\/strong>
\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\"Selina<\/p>\n

Question 3(iv).<\/strong>
\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\"Selina<\/p>\n

Question 4.<\/strong>
\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\"Selina<\/p>\n

Question 5.<\/strong>
\nSolve:
\n(i) x4<\/sup>\u00a0– 2x2<\/sup>\u00a0– 3 =0
\n(ii) x4<\/sup>\u00a0– 10x2<\/sup>\u00a0+9 =0
\nSolution:<\/strong>
\n\"Selina<\/p>\n

Question 6.<\/strong>
\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\"Selina<\/p>\n

Question 7.<\/strong>
\n\"Selina
\nSolution:<\/strong>
\n\"Selina
\n\"Selina
\n\"Selina<\/p>\n

Question 8.<\/strong>
\n\"Selina
\nSolution:<\/strong>
\n\"Selina<\/p>\n

Question 9.<\/strong>
\nSolve the following equation and give your answer correct to 3 significant figures:
\n5x\u00b2 – 3x – 4 = 0
\nSolution:<\/strong>
\n\"Selina<\/p>\n

Question 10.<\/strong>
\nSolve for x using the quadratic formula. Write your answer correct to two significant figures.
\n(x – 1)2<\/sup>\u00a0– 3x + 4 = 0
\nSolution:<\/strong>
\n\"Selina<\/p>\n

Question 11.<\/strong>
\nSolve the quadratic equation x\u00b2\u00a0– 3 (x+3) = 0; Give your answer correct to two significant figures.
\nSolution:
\n\"Selina
\n<\/strong><\/p>\n

Quadratic Equations Exercise 5E – Selina Concise Mathematics Class 10 ICSE Solutions<\/h3>\n

Question 1.<\/strong>
\n\"Selina
\nSolution:<\/strong>
\n\"Selina<\/p>\n

Question 2.<\/strong>
\nSolve: (2x+3)2<\/sup>=81
\nSolution:<\/strong>
\n\"Selina<\/p>\n

Question 3.<\/strong>
\nSolve: a\u00b2x\u00b2 – b\u00b2 = 0
\nSolution:<\/strong>
\n\"Selina<\/p>\n

Question 4.<\/strong>
\n\"Selina
\nSolution:<\/strong>
\n\"Selina<\/p>\n

Question 5.<\/strong>
\n\"Selina
\nSolution:<\/strong>
\n\"Selina<\/p>\n

Question 6.<\/strong>
\nSolve: 2x4<\/sup>\u00a0– 5x\u00b2 + 3 = 0
\nSolution:<\/strong>
\n\"Selina<\/p>\n

Question 7.<\/strong>
\nSolve: x4<\/sup>\u00a0– 2x\u00b2 – 3 = 0.
\nSolution:<\/strong>
\n\"Selina<\/p>\n

Question 8.<\/strong>
\n\"Selina
\nSolution:<\/strong>
\n\"Selina<\/p>\n

Question 9.<\/strong>
\n\"Selina
\nSolution:<\/strong>
\n\"Selina<\/p>\n

Question 10.<\/strong>
\n\"Selina
\nSolution:<\/strong>
\n\"Selina<\/p>\n

Question 11.<\/strong>
\nSolve : (x\u00b2\u00a0+ 5x + 4)(x\u00b2 + 5x + 6) = 120
\nSolution:<\/strong>
\n\"Selina<\/p>\n

Question 12.<\/strong>
\nSolve each of the following equations, giving answer upto two decimal places.
\n(i) x2<\/sup>\u00a0– 5x -10=0 (ii) 3x2<\/sup>\u00a0– x – 7 =0
\nSolution:<\/strong>
\n\"Selina<\/p>\n

Question 13.<\/strong>
\n\"Selina
\nSolution:<\/strong>
\n\"Selina<\/p>\n

Question 14.<\/strong>
\nSolve :
\n(i) x2<\/sup>\u00a0– 11x – 12 =0; when x \u2208\u00a0N
\n(ii) x2<\/sup>\u00a0– 4x – 12 =0; when x \u2208\u00a0I
\n(iii) 2x2<\/sup>\u00a0– 9x + 10 =0; when x\u00a0\u2208 Q
\nSolution:<\/strong>
\n\"Selina<\/p>\n

Question 15.<\/strong>
\nSolve : (a + b)\u00b2x\u00b2 – (a + b)x – 6 = 0; a + b\u00a0\u2260 0
\nSolution:<\/strong>
\n\"Selina<\/p>\n

Question 16.<\/strong>
\n\"Selina
\nSolution:<\/strong>
\n\"Selina<\/p>\n

Question 17.<\/strong>
\n\"Selina
\nSolution:<\/strong>
\n\"Selina<\/p>\n

Question 18.<\/strong>
\n\"Selina
\nSolution:<\/strong>
\n\"Selina
\n\"Selina<\/p>\n

Question 19.<\/strong>
\n\"Selina
\nSolution:<\/strong>
\n\"Selina<\/p>\n

Question 20.<\/strong>
\nWithout solving the following quadratic equation, find the value of ‘m’ for which the given equation has real and equal roots.
\nx\u00b2\u00a0+ 2(m – 1)x + (m + 5) = 0
\nSolution:<\/strong>
\n\"Selina<\/p>\n

Quadratic Equations Exercise 5F – Selina Concise Mathematics Class 10 ICSE Solutions<\/h3>\n

Solution 1(i)
\nGiven: (x + 5)(x – 5)=24
\n\u21d2 x2<\/sup> – 52<\/sup> = 24\u00a0 \u00a0…. since (a – b)(a + b) = a2<\/sup> – b2<\/sup>
\n\u21d2 x2<\/sup> – 25 = 24
\n\u21d2 x2<\/sup> = 49
\n\u21d2 x = \u00b1 7<\/p>\n

Solution 1(ii)
\nGiven: 3x2<\/sup> – 2\\(\\sqrt{6}\\)x + 2 = 0
\n\"Selina
\nSolution 1(iii)
\nGiven: 3\\(\\sqrt{2}\\)x2<\/sup> – 5x –\u00a0\\(\\sqrt{26}\\) = 0
\n\"Selina
\nQuestion 2.
\nOne root of the quadratic equation\u00a08x2<\/sup> + mx + 15\u00a0is 3\/4<\/span>. Find the value of m. Also, find the other root of the equation.<\/span>
\nSolution:
\nGiven quadratic equation is\u00a0 8x2<\/sup> + mx + 15 = 0\u00a0 \u00a0\u2026. (i)
\nOne of the roots of (i) is \\(\\frac{3}{4}\\), so it satisfies (i)
\n\"Selina
\nSo, the equation (i) becomes 8x2<\/sup> – 26x + 15 = 0
\n\u21d2 8x2<\/sup> – 20x – 6x + 15 = 0
\n\u21d2 4x(2x – 5) -3(2x – 5) = 0
\n\u21d2 (4x – 3)(2x – 5) = 0
\n\u21d2 x = \\(\\frac{3}{4}\\) or x = \\(\\frac{5}{2}\\)
\n\u21d2 x = \\(\\frac{3}{4}, \\frac{5}{2}\\)
\nHence, the other root is\u00a0\\(\\frac{5}{2}\\)<\/p>\n

Question 3.
\nOne root of the quadratic equation\u00a0x2<\/sup> – 3x – 2ax – 6a = 0\u00a0is -3, find its other root.<\/span>
\nSolution:
\nGiven quadratic equation is \u2026. (i)
\nOne of the roots of (i) is -3, so it satisfies (i)
\n\u21d2 x2<\/sup> – 3x – 2ax – 6a = 0
\n\u21d2 x(x + 3) – 2a(x + 3) = 0
\n\u21d2 (x – 2a)(x + 3) = 0
\n\u21d2 x = -3, 2a
\nHence, the other root is 2a.<\/p>\n

Question 4.
\nIf\u00a0p – 15 = 0 and\u00a02x2<\/sup> + 15x + 15 = 0;find the values of x.<\/span>
\nSolution:
\nGiven i.e p – 15 = 0 i.e. p = 15
\nSo, the given quadratic equation becomes
\n2x2<\/sup> + 15x + 15 = 0
\n\u21d2 2x + 10x + 5x + 15 = 0
\n\u21d2 2x(x + 5) + 5(x + 5)
\n\u21d2 (2x + 5)(x + 5) = 0
\n\u21d2 x = -5, \\(-\\frac{5}{2}\\)
\nHence, the values of x are -5 and \\(-\\frac{5}{2}\\)<\/p>\n

Question 5.
\nFind the solution of the equation 2x2<\/sup> -mx – 25n = 0; if\u00a0m + 5 = 0\u00a0and<\/span><\/span>\u00a0n – 1 = 0.<\/span>
\nSolution:
\nGiven quadratic equation is 2x2<\/sup> -mx – 25n = 0 \u2026.. (i)
\nAlso, given and m + 5 = 0 and n – 1 = 0
\n\u21d2 m = -5 and n = 1
\nSo, the equation (i) becomes
\n2x2<\/sup> + 5x + 25 = 0
\n\u21d2 2x + 10x – 5x – 25 = 0
\n\u21d2 2x(x + 5) -5(x + 5) = 0
\n\u21d2 (x + 5)(2x – 5) = 0
\n\u21d2 x = -5, \\(\\frac{5}{2}\\)
\nHence, the solution of given quadratic equation are x and \\(\\frac{5}{2}\\)<\/p>\n

Question 6.
\nIf m and n are roots of the equation\u00a0\\(\\frac{1}{x}-\\frac{1}{x-2}=3\\) where x \u2260 0 and x \u2260 2; find m \u00d7 n.<\/span>
\nSolution:
\nGiven quadratic equation is \\(\\frac{1}{x}-\\frac{1}{x-2}=3\\)
\n\"Selina
\nSince, m and n are roots of the equation, we have
\n\"Selina<\/p>\n

Question 7.
\nSolve, using formula :
\nx2<\/sup> + x – (a + 2)(a + 1) = 0
\nSolution:
\nGiven quadratic equation is x2<\/sup> + x – (a + 2)(a + 1) = 0
\nUsing quadratic formula,
\n\"Selina<\/p>\n

Question 8.
\nSolve the quadratic equation\u00a08x2<\/sup> – 14x + 3 = 0
\n(i) When x \u2208 I (integers)
\n(ii) When\u00a0x \u2208 Q (rational numbers)
\nSolution:
\nGiven quadratic equation is 8x2<\/sup> – 14x + 3 = 0
\n\u21d2 8x2<\/sup> – 12x – 2x + 3 = 0
\n\u21d2 4x(2x – 3) – (2x – 3) = 0
\n\u21d2 (4x – 1)(2x – 3) = 0
\n\u21d2 x = \\(\\frac{3}{2}\\) or x = \\(\\frac{1}{4}\\)
\n(i) When x \u03f5 I, the equation 8x2<\/sup> – 14x + 3 = 0 has no roots
\n(ii) When x \u03f5 Q the roots of 8x2<\/sup> – 14x + 3 = 0 are
\nx = \\(\\frac{3}{2}\\) x = \\(\\frac{1}{4}\\)<\/p>\n

Question 9.
\nFind the value of m for which the equation\u00a0(m + 4 )2<\/sup> + (m + 1)x + 1 = 0\u00a0has real and equal roots.
\nSolution:
\nGiven quadratic equation is (m + 4 )2<\/sup> + (m + 1)x + 1 = 0
\nThe quadratic equation has real and equal roots if its discriminant is zero.
\n\u21d2 D = b2<\/sup> – 4ac = 0
\n\u21d2 (m + 1)2<\/sup> -4(m + 4)(1) = 0
\n\u21d2 m2<\/sup> + 2m + 1 – 4m – 16 = 0
\n\u21d2 m2<\/sup> – 2m – 15 = 0
\n\u21d2 m2<\/sup> – 5m + 3m – 15 = 0
\n\u21d2 m(m – 5) +3(m =5) = 0
\n\u21d2 (m – 5)(m + 3) = 0
\n\u21d2 m = 5 or m = -3<\/p>\n

Question 10.
\nFind the values of m for which equation\u00a03x2<\/sup> + mx + 2 = 0\u00a0has equal roots. Also, find the roots of the given equation.
\nSolution:
\nGiven quadratic equation is 3x2<\/sup> + mx + 2 = 0 \u2026. (i)
\nThe quadratic equation has equal roots if its discriminant is zero
\n\u21d2 D = b2<\/sup> – 4ac = 0
\n\u21d2 m2<\/sup> – 4(2)(3) = 0
\n\u21d2 m2<\/sup> = 24
\n\u21d2 m = \\(\\pm 2 \\sqrt{6}\\)
\nWhen m = \\(2 \\sqrt{6}\\), equation (i) becomes
\n\"Selina
\nWhen m = \\(-2 \\sqrt{6}\\), equation (i) becomes
\n\"Selina
\n\u2234 x= \\(-\\frac{\\sqrt{6}}{3}, \\frac{\\sqrt{6}}{3}\\)<\/p>\n

Question 11.
\nFind the value of k for which equation\u00a04x2<\/sup> + 8x – k = 0\u00a0has real roots.<\/span>
\nSolution:
\nGiven quadratic equation is 4x2<\/sup> + 8x – k = 0 \u2026. (i)
\nThe quadratic equation has real roots if its discriminant is greater than or equal to zero
\n\u21d2 D = b2<\/sup> – 4ac \u2265 0
\n\u21d2 82<\/sup> – 4(4)(-k) \u2265 0
\n\u21d2 64 + 16k \u2265 0
\n\u21d2 16k \u2265 -64
\n\u21d2 k \u2265 -4
\nHence, the given quadratic equation has real roots for k \u2265 -4<\/p>\n

Question 12.
\nFind, using quadratic formula, the roots of the following quadratic equations, if they exist
\n(i) 3x2<\/sup> – 5x + 2 = 0
\n(ii) x2<\/sup> + 4x + 5 = 0
\nSolution:
\n(i) Given quadratic equation is 3x2<\/sup> – 5x + 2 = 0
\nD = b2<\/sup> – 4ac = (-5)2<\/sup> – 4(3)(2) = 25 – 24 = 1
\nSince D > 0, the roots of the given quadratic equation are real and distinct.
\nUsing quadratic formula, we have
\n\"Selina
\n\u21d2 x = 1 or x = \\(\\frac{2}{3}\\)<\/p>\n

(ii) Given quadratic equation is x2<\/sup> + 4x + 5 = 0
\nD = b2<\/sup> – 4ac = (4)2<\/sup> – 4(1)(5) = 16 – 20 = – 4
\nSince D < 0, the roots of the given quadratic equation does not exist.<\/p>\n

Solution 13:
\n(i) Given quadratic equation is \\(\\frac{1}{18-x}-\\frac{1}{18+x}=\\frac{1}{24}\\)
\n\"Selina
\n\u21d2 48x = 324 – x2<\/sup>
\n\u21d2 x2<\/sup> + 48x – 324 = 0
\n\u21d2 x2<\/sup> + 54x – 6x – 324 = 0
\n\u21d2 x(x + 54) -6(x + 54) = 0
\n\u21d2 (x + 54)(x – 6) = 0
\n\u21d2 x = -54 or x = 6
\nBut as x > 0, so x can’t be negative.
\nHence, x = 6.
\n(ii) Given quadratic equation is \\((x-10)\\left(\\frac{1200}{x}+2\\right)=1260\\)
\n\u21d2 (x – 10)\\(\\left(\\frac{1200+2 x}{x}\\right)\\) = 1260
\n\u21d2 (x – 10)(1200 + 2x) = 1260x
\n\u21d2 1200x + 2x2<\/sup> – 12000 – 20x = 1260x
\n\u21d2 2x2<\/sup> – 12000 – 80x = 0
\n\u21d2 x2<\/sup> – 40x – 6000 = 0
\n\u21d2 x2<\/sup> – 100x + 60x – 6000 = 0
\n\u21d2 (x – 100)(x – 60) = 0
\n\u21d2 x = 100 or x = -60
\nBut as x < 0, so x can’t be positive.
\nHence, x = -60.<\/p>\n

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