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3. 3. Quadratic Equations Quadratic Equations MODULE - I p r∴ 2 The roots of the quadratic equation ax + bx + c = 0 are q , s AlgebraExample 2.1 Using factorization method, solve the following quadratic equation : 6x 2 + 5x − 6 = 0Solution: The given quadratic equation is Notes 6x + 5x − 6 = 0 2 ... (i) Splitting the middle term, we have 6x 2 + 9x – 4x − 6 = 0 or, 3x (2x + 3) –2 (2x + 3) = 0 or, (2x + 3)(3x – 2) = 0 3 ∴ Either 2x + 3 = 0 ⇒ x = − 2 2 or, 3x – 2 = 0 ⇒ x = 3 3 2 ∴ Two roots of the given quadratic equation are − , 2 3Example 2.2 Using factorization method, solve the following quadratic equation: 3 2 x + 7x − 3 2 = 0 2Solution: Splitting the middle term, we have 3 2 x + 9x – 2x − 3 2 = 0 2 or, 3x ( 2 x + 3) − 2 ( 2 x + 3) = 0 or, ( 2 x + 3)(3x − 2 ) = 0 3 ∴ Either 2x+3=0 ⇒ x= − 2 2 or, 3x − 2 = 0 ⇒ x= 3 3 2 ∴ Two roots of the given quadratic equation are − , 2 3MATHEMATICS 41
4. 4. Quadratic EquationsMODULE - I Example 2.3 Using factorization method, solve the following quadratic equation: Algebra (a + b)2 x2 + 6 (a2 − b2) x + 9 (a – b)2 = 0 Solution: The given quadratic equation is (a + b)2 x2 + 6 (a2 − b2) x + 9 (a – b)2 = 0 Notes Splitting the middle term, we have (a + b)2 x2 + 3(a2 − b2) x + 3(a2 – b2) x + 9 (a – b)2 = 0 or, (a + b) x {(a + b) x + 3 (a − b) } + 3 (a – b) {(a +b) x + 3 (a – b) } = 0 or, {(a + b) x + 3 (a − b) } {(a + b) x + 3 (a – b) } = 0 − 3(a − b) 3 (b − a) ∴ either (a + b) x + 3 (a – b) =0 ⇒ x = = a+b a+b − 3(a − b) 3 (b − a) or, (a + b) x + 3 (a – b) =0 ⇒ x = = a+b a+b 3 (b − a) 3 (b − a) The equal roots of the given quadratic equation are , a+b a+b Alternative Method The given quadratic equation is (a + b)2 x2 + 6(a2 − b2) x + 9(a – b)2 = 0 This can be rewritten as {(a + b) x}2 + 2 .(a + b)x . 3 (a – b) + {3(a – b)}2 = 0 or, { (a + b)x + 3(a – b) }2 = 0 3(a − b) 3 (b − a) or, x= − = a+b a+b 3 (b − a) 3 (b − a) ∴ The quadratic equation have equal roots , a+b a+b CHECK YOUR PROGRESS 2.1 1. Solve each of the following quadratic equations by factorization method: 2 (i) 3 x + 10x + 8 3 =0 (ii) x2 – 2ax + a2 – b = 0  ab c  (iii) x2 +  −  x – 1 = 0 (iv) x2 – 4 2x+6=0  c ab  42 MATHEMATICS
5. 5. Quadratic Equations Quadratic Equations MODULE - I 2.3 SOLVING QUADRATIC EQUATION BY QUADRATIC Algebra FORMULARecall the solution of a standard quadratic equationa x2 + bx + c = 0, a ≠ 0 by the “Method of Completing Squares”Roots of the above quadratic equation are given by Notes − b + b 2 − 4ac − b − b 2 − 4acx1= and x 2 = 2a 2a −b+ D −b− D = , = 2a 2awhere D = b2 – 4ac is called the discriminant of the quadratic equation. For a quadratic equation ax 2 + bx + c = 0, a ≠ 0 if (i) D>0, the equation will have two real and unequal roots (ii) D=0, the equation will have two real and equal roots and both roots are b equal to − 2a (iii) D<0, the equation will have two conjugate complex (imaginary) roots.Example 2.4 Examine the nature of roots in each of the following quadratic equations and also verify them by formula.(i) x2 + 9 x +10 = 0 (ii) 9 y2 − 6 2 y + 2 = 0 2 t − 3t + 3 2 = 0 2(iii)Solution:(i) The given quadratic equation is x2 + 9 x + 10 = 0Here, a = 1, b = 9 and c= 10∴ D = b2 – 4ac = 81 – 4.1.10 = 41>0.∴ The equation will have two real and unequal rootsMATHEMATICS 43
6. 6. Quadratic EquationsMODULE - I Verification: By quadratic formula, we have Algebra − 9 ± 41 x= 2 − 9 + 41 − 9 − 41 Notes ∴ The two roots are , which are real and unequal. 2 2 (ii) The given quadratic equation is 9 y2 – 6 2 y + 2 = 0 Here, D = b2 – 4 ac = ( − 6 2 )2 – 4.9.2 = 72 – 72 = 0 ∴ The equation will have two real and equal roots. Verification: By quadratic formula, we have 6 2± 0 2 y= = 2.9 3 2 2 ∴ The two equal roots are , . 3 3 2 t − 3t + 3 2 = 0 2 (iii) The given quadratic equation is Here, D = ( − 3)2 − 4. 2 .3 2 = − 15< 0 ∴ The equation will have two conjugate complex roots. Verification: By quadratic formula, we have 3 ± −15 t= 2 2 3 ± 15i = , where i = − 1 2 2 3+ 15 i 3− 15 i ∴ Two conjugate complex roots are , 2 2 2 2 44 MATHEMATICS