Quadratic equations

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  • 1. Quadratic Equations Quadratic Equations MODULE - I Algebra 2 Notes QUADRATIC EQUATIONSRecall that an algebraic equation of the second degree is written in general form asax 2 + bx + c = 0, a ≠ 0It is called a quadratic equation in x. The coefficient ‘a’ is the first or leading coefficient, ‘b’is the second or middle coefficient and ‘c’ is the constant term (or third coefficient). 5 1For example, 7x² + 2x + 5 = 0, x² + x + 1 = 0, 2 2 13x² − x = 0, x² + = 0, 2 x² + 7x = 0, are all quadratic equations. 2In this lesson we will discuss how to solve quadratic equations with real and complexcoefficients and establish relation between roots and coefficients. We will also find cuberoots of unity and use these in solving problems. OBJECTIVESAfter studying this lesson, you will be able to:• solve a quadratic equation with real coefficients by factorization and by using quadratic formula;• find relationship between roots and coefficients;• form a quadratic equation when roots are given; and• find cube roots of unity. EXPECTED BACKGROUND KNOWLEDGE• Real numbers• Quadratic Equations with real coefficients.MATHEMATICS 39
  • 2. Quadratic EquationsMODULE - I 2.1 ROOTS OF A QUADRATIC EQUATION Algebra The value which when substituted for the variable in an equation, satisfies it, is called a root (or solution) of the equation. If  be one of the roots of the quadratic equation Notes then, ax 2 + bx + c = 0, a ≠ 0 ... (i) a 2 + b + c = 0 In other words, x −  is a factor of the quadratic equation (i) In particular, consider a quadratic equation x2 + x − 6 = 0 ...(ii) If we substitute x = 2 in (ii), we get L.H.S = 22 + 2 – 6 = 0 ∴ L.H.S = R.H.S. Again put x = − 3 in (ii), we get L.H.S. = ( − 3)2 –3 –6 = 0 ∴ L.H.S = R.H.S. Again put x = − 1 in (ii) ,we get L.H.S = ( − 1)2 + ( − 1) – 6 = –6 ≠ 0 = R.H.S. ∴ x = 2 and x = − 3 are the only values of x which satisfy the quadratic equation (ii) There are no other values which satisfy (ii) ∴ x = 2, x = − 3 are the only two roots of the quadratic equation (ii) Note: If  ,  be two roots of the quadratic equation ax 2 + bx + c = 0, a ≠ 0 ...(A) then (x −  ) and (x −  ) will be the factors of (A). The given quadratic equation can be written in terms of these factors as (x −  ) (x −  ) = 0 2.2 SOLVING QUADRATIC EQUATION BY FACTORIZATION Recall that you have learnt how to factorize quadratic polynomial of the form p ( x ) = ax 2 + bx + c , a ≠ 0, by splitting the middle term and taking the common factors. Same method can be applied while solving quadratic equation by factorization. p r If x − q and x − s are two factors of the quadratic equation p r ax2 + bx + c = 0 ,a ≠ 0 then ( x − q )( x − s ) = 0 p r ∴ either x = q or, x = s 40 MATHEMATICS
  • 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. 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. 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. 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
  • 7. Quadratic Equations Quadratic EquationsExample 2.5 Prove that the quadratic equation x2 + py – 1 = 0 has real and distinct roots MODULE - I for all real values of p. AlgebraSolution: Here, D = p2 + 4 which is always positive for all real values of p.∴ The quadratic equation will have real and distinct roots for all real values of p.Example 2.6 For what value of k the quadratice equation Notes (4k+ 1) x2 + (k + 1) x + 1 = 0 will have equal roots ?Solution: The given quadratic equation is (4k + 1)x2 + (k + 1) x + 1 = 0Here, D = (k + 1)2 – 4.(4k + 1).1 For equal roots, D = 0∴ (k + 1)2 – 4 (4k + 1) = 0⇒ k 2 – 14k –3 = 0 14 ± 196 +12∴ k= 2 14 ± 208or k= 2 = 7 ± 2 13 or 7 + 2 13 , 7 – 2 13which are the required values of k.Example 2.7 Prove that the roots of the equation x2 (a2+ b2) + 2x (ac+ bd) + (c2+ d2) = 0 are imaginary. But if ad = bc, roots are real and equal.Solution: The given equation is x2 (a2 + b2) + 2x (ac + bd) + (c2 + d2) = 0Discriminant = 4 (ac + bd)2 – 4 (a2 + b2) (c2 + d2) = 8 abcd – 4(a2 d2 + b2 c2) = − 4 ( − 2 abcd + a2 d2 + b2 c2) = − 4 (ad –bc) 2 <0 for all a, b, c, d∴ The roots of the given equation are imaginary. For real and equal roots, discriminant is equal to zero.⇒ − 4 (ad − bc)2 = 0or, ad = bc Hence, if ad=bc, the roots are real and equal.MATHEMATICS 45
  • 8. Quadratic EquationsMODULE - I Algebra CHECK YOUR PROGRESS 2.2 1. Solve each of the following quadratic equation by quadratic formula: Notes (i) 2x2 – 3 x + 3 = 0 (ii) − x 2 + 2x − 1 = 0 (iii) −4 x 2 + 5 x − 3 = 0 (iv) 3x 2 + 2 x + 5 = 0 2. For what values of k will the equation y2 – 2 (1 + 2k) y + 3 + 2k = 0 have equal roots ? 3. Show that the roots of the equation (x − a) (x − b) + (x − b) (x − c) + (x − c) (x − a) = 0 are always real and they can not be equal unless a = b = c. 2.4 RELATION BETWEEN ROOTS AND COEFFICIENTS OF A QUADRATIC EQUATION You have learnt that, the roots of a quadratic equation ax2 + bx + c = 0, a ≠ 0 − b + b 2 − 4 ac − b − b 2 − 4 ac are and 2a 2a − b + b 2 − 4 ac − b − b 2 − 4 ac Let  = ...(i) and  = ... (ii) 2a 2a Adding (i) and (ii), we have − 2b −b  + = = 2a a coefficient of x b ∴ Sum of the roots = – coefficient of x 2 = − a ... (iii) + b 2 − (b 2 − 4 ac)   = 4a 2 4 ac = 4a 2 c = a 46 MATHEMATICS
  • 9. Quadratic Equations Quadratic Equations MODULE - I∴ constant term c Product of the roots = coefficient of x 2 = a ...(iv) Algebra(iii) and (iv) are the required relationships between roots and coefficients of a given quadraticequation. These relationships helps to find out a quadratic equation when two roots aregiven. NotesExample 2.8 If,  ,  are the roots of the equation 3x2 – 5x + 9 = 0 find the value of: 1 1 (a)  2 +  2 (b) + 2 2Solution: (a) It is given that  ,  are the roots of the quadratic equation 3x2 – 5x +9 = 0. 5∴  + = ... (i) 3 9and  = =3 ... (ii) 3 2 + 2 = (  +  ) − 2 2Now, 2  5 =   − 2.3 [By (i) and (ii)]  3 29 =− 9 29∴ The required value is − 9 1 1 2 +2(b) Now, + 2 = 2  2 2 − 29 = 9 [By (i) and (ii)] 9 29 =− 81MATHEMATICS 47
  • 10. Quadratic EquationsMODULE - I Example 2.9 If  ,  are the roots of the equation 3y2 + 4y + 1 = 0, form a quadratic Algebra equation whose roots are  2,  2 Solution: It is given that  ,  are two roots of the quadratic equation 3y2 + 4y + 1 = 0. ∴ Sum of the roots Notes coefficient of y i.e.,  +  = − coefficient of y 2 4 =− ... (i) 3 cons tan t term Product of the roots i.e.,   = coefficient of y 2 1 = ... (ii) 3 Now,  2 +  2 = (  +  )2 – 2   2  4 1 =  −  − 2. [ By (i) and (ii)]  3 3 16 2 = − 9 3 10 = 9 1 and  2  2 = (   )2 = [By (i) ] 9 ∴ The required quadratic equation is y2 – (  2 +  2)y +  2  2 = 0 10 1 or, y2 − y+ =0 9 9 or, 9y2 – 10y + 1 = 0 Example 2.10 If one root of the equation ax 2 + bx + c = 0, a ≠ 0 be the square of the other, prove that b3 + ac2 + a2c = 3abc Solution: Let  ,  2 be two roots of the equation ax2 + bx + c = 0. b ∴  + 2= − ... (i) a 48 MATHEMATICS
  • 11. Quadratic Equations Quadratic Equations MODULE - Iand  . 2 = c Algebra a ci.e., 3= . ... (ii) a NotesFrom (i) we have b  (  + 1) = − a 3  b b3or, { ( + 1)} 3 = −  = − 3  a a b3or,  3 (  3 + 3  2 + 3  +1) = − a3 c c  b  3or,  + 3  −  +1 = − b3 ... [ By (i) and (ii)] a a  a  a c2 3 bc c b3or, – + =– 3 a2 a2 a aor, ac2 – 3abc + a2c = − b3or, b3 + ac2 + a2c = 3abcwhich is the required result.Example 2.11 Find the condition that the roots of the equation ax2 + bx + c = 0 are in the ratio m : nSolution: Let m  and n  be the roots of the equation ax2 +bx + c = 0 bNow, m  + n  = – ... (i) a cand mn  2 = ... (ii) a bFrom (i) we have,  (m + n ) = – a b2or,  (m + n)2 = 2 a2MATHEMATICS 49
  • 12. Quadratic EquationsMODULE - I c b2 Algebra or, a (m + n)2 = mn 2 a [ By (ii)] or, ac (m+n)2 = mn b2 which is the required condition Notes CHECK YOUR PROGRESS 2.3 1. If  ,  are the roots of the equation ay2 + by + c = 0 then find the value of : 1 1 1 1 (i) + 2 (ii) + 4  2 4 2. If  ,  are the roots of the equation 5x2 – 6x + 3 = 0, form a quadratic equation whose roots are: (i)  2,  2 (ii)  3  ,   3 3. If the roots of the equation ay2 + by + c = 0 be in the ratio 3:4, prove that 12b2 = 49 ac 4. Find the condition that one root of the quadratic equation px2 – qx + p = 0 may be 1 more than the other. 2.5 SOLUTION OF A QUADRATIC EQUATION WHEN D < 0 Let us consider the following quadratic equation: (a) Solve for t : t2 + 3t + 4 = 0 − 3 ± 9 −16 −3 ± −7 ∴ t= = 2 2 Here, D= –7 < 0 −3 + −7 −3− − 7 ∴ The roots are and 2 2 − 3 + 7 i −3 − 7 i or, , 2 2 Thus, the roots are complex and conjugate. 50 MATHEMATICS
  • 13. Quadratic Equations Quadratic Equations(b) Solve for y : MODULE - I Algebra − 3y2 + 5 y – 2 = 0 − 5 ± 5 − 4(−3).(−2)∴ y= 2(−3) Notes − 5 ± −19or y= −6Here, D = − 19 < 0 − 5 + 19i 5 − 19i∴ The roots are ,− −6 −6Here, also roots are complex and conjugate. From the above examples , we can make thefollowing conclusions: (i) D < 0 in both the cases (ii) Roots are complex and conjugate to each other.Is it always true that complex roots occur in conjugate pairs ?Let us form a quadratic equation whose roots are 2 + 3i and 4 – 5iThe equation will be {x – (2 + 3i)} {x – (4 – 5i)} = 0or, x2 – (2 + 3i)x – (4 – 5i)x + (2 + 3i) (4 – 5i) = 0or, x2 + (–6 + 2i)x + 23 + 2i = 0which is an equation with complex coefficients. Note : If the quadratic equation has two complex roots, which are not conjugate of each other, the quadratic equation is an equation with complex coefficients. 2.6 CUBE ROOTS OF UNITYLet us consider an equation of degree 3 or more. Any equation of degree 3 can be expressedas a product of a linear and quadratic equation.The simplest situation that comes for our consideration is x³ − 1 = 0 ...(i)∴ x³ − 1 = (x − 1) (x ² + x + 1) = 0either x − 1 = 0 or, x²+x+1=0MATHEMATICS 51
  • 14. Quadratic EquationsMODULE - I Algebra −1± 3i or, x=1 or, x= 2 1 3i 1 3i ∴ Roots are 1, − + and − − 2 2 2 2 Notes These are called cube roots of unity. Do you see any relationship between two non-real roots of unity obtained above ? Let us try to find the relationship between them 1 3i Let w= − + 2 2 Squaring both sides, we have 2  1 3 i = − +  2  2  w²   1 = (1+ 3 i 2 − 2 3 i) 4 1 = (1− 3 − 2 3 i) 4 −2−2 3 i = 4 = ( − 2 1+ 3 i ) = ( − 1+ 3 i ) 4 2 1 3i ∴ w² = − − = other complex root. 2 2 ∴ We denote cube roots of unity as 1, w, w² 52 MATHEMATICS
  • 15. Quadratic Equations Quadratic Equations MODULE - I 1Note: If we take w = − − 3i , we can verify, that in this case Algebra 2 2 1 3i also w2 = − + = other complex root. 2 2 NotesThus, square of one complex root is same as the other complex root.Again, sum of roots i.e., 1+ w + w²  1 3 i  1 3 i 1 + − +  2  + − −  2   2 2 =     1 1= 1− − 2 2= 1−1= 0∴ 1 + w + w² = 0Sum of cube roots of unity is zero.Product of cube roots of 1i.e., 1.w.w² = w³ = 1 (since ,w is a root the equation x3 − 1 = 0) We can conclude that , if 1, w and w2 are cube roots of unity then (i) square of one complex root is same as the other complex root i.e., w2 = w (ii) 1+ w + w2 = 0 (iii) w3 = 1Let us now consider another equation x³ + 1 = 0 ... (ii)or (x + 1) (x ² − x + 1) = 0Either, x + 1 = 0 or, x² −x+1=0 1± 1− 4or, x = −1 or, x= 2 1± −3⇒ x = −1 or, x= 2MATHEMATICS 53
  • 16. Quadratic EquationsMODULE - I Algebra ∴ Roots are − 1, 1 − 3i 1 and + 3i which can also be written as − 1 , − w and − w2 2 2 2 2 Therefore, cube roots of − 1 are − 1 , − w and − w2 In general, roots of any cubic equation of the form x 3 = ± a3 would be Notes ± a , ± aw and ± aw2 Example 2.12 If 1, w and w² are cube roots of unity, prove that (a) 1 + w2 + w7 = 0 (b) (1 − w + w2) (1 + w – w2) = 4 (c) (1 + w)3 − (1 + w2)3 = 0 (d) (1 − w + w2)3 = − 8 and (1 + w − w2)3 = − 8 Solution: (a) 1 + w2 + w7 = 0 L.H.S = 1 + w2 + (w3)2. w = 1 + w2 + w [ since w3 =1] =0 [ since 1+ w + w2 = 0] = R.H.S ∴ L.H.S = R.H.S (b) (1 − w + w2) (1 + w – w2) = 4 L.H.S = (1 − w + w2) (1 + w – w2) = (1 + w2 − w) (1 + w – w2) (since 1 + w2 = − w and 1 + w = − w2) = ( − w − w) ( − w2 –w2) = ( − 2w) ( − 2w2) = 4w3 = 4.1 = 4 =R.H.S ∴ L.H.S = R.H.S 54 MATHEMATICS
  • 17. Quadratic Equations Quadratic Equations(c) (1 + w)3 − (1 + w2)3 = 0 MODULE - I Algebra L.H.S = (1 + w)3 − (1 + w2)3 = ( − w2)3 − ( − w)3 (∴ 1 + w = − w2 = − w6 + w3 and 1 + w2 = − w) = − (w3)2 + 1 Notes = − (1)2 + 1 = 0 =R.H.S∴ L.H.S = R.H.S(d) (1 − w + w2)3 = − 8 and (1 + w − w2)3 = − 8Case I : L.H.S = (1 − w + w2)3 = (1 + w2 − w)3 = ( − w − w)3 = ( − 2w)3 = − 8w3 = − 8 = R.H.S ∴ L.H.S = R.H.SCase II : L.H.S = (1 + w − w²)³ = ( − w2 – w2)3 = ( − 2w2)3 = − 8w6 = − 8(w3)2 = − 8 = R.H.S ∴ L.H.S = R.H.S CHECK YOUR PROGRESS 2.41. Solve each of the following cubic equations: (i) x3 = 27 (ii) x3 = − 27 (iii) x3 = 64 (iv) x3 = − 64MATHEMATICS 55
  • 18. Quadratic EquationsMODULE - I 2. If 1, w, w2 are cube roots of unity, show that Algebra (i) (1 + w) (1 + w2) (1 + w4) (1 + w8) = 1 (ii) (1 − w) (1 − w2) (1 – w4) (1 – w5) = 9 Notes (iii) (1 + w)4 + (1 + w2)4 = − 1 (iv) (1 + w3)3 = 8 (v) (1 – w + w2)6 = (1 + w – w2)6 = 64 (vi) (1 + w)16 + w = (1 + w2)16 + w2 = − 1 LET US SUM UP • Roots of the quadratic equation ax2 + bx + c = 0 are complex and conjugate of each other, when D < 0. • If  ,  be the roots of the quadratic equation b c ax2 + bx + c = 0 then  +  = − and   = a a • Cube roots of unity are 1, w, w2 1 3i 1 3i where w = − + and w 2 = − − 2 2 2 2 • Sum of cube roots of unity is zero i.e., 1+ w + w2 =0 • Product of cube roots of unity is 1 i.e., w3 = 1 • Complex roots w and w2 are conjugate to each other. • In general roots of any cubic equation of the form x 3 = ± a3 would be ± a , ± aw and ± aw2 56 MATHEMATICS
  • 19. Quadratic Equations Quadratic Equations MODULE - I Algebra SUPPORTIVE WEB SITES http://www.wikipedia.org http://mathworld.wolfram.com Notes TERMINAL EXERCISE1. Show that the roots of the equation 2(a2 + b2 )x2 + 2(a + b)x + 1=0 are imaginary,when a ≠ b2. Show that the roots of the equation bx2 + ( b – c)x = c + a –b are always real if those of ax2 + b( 2x + 1) = 0 are imaginary.3. If  ,  be the roots of the equation 2x2 – 6x + 5 = 0 , find the equation whose roots are:   1 1 1 1 (i)  , (ii)  +  ,  + (iii)  2 +  2 , + 2   24. If 1, w and w2 are cube roots of unity , prove that (a) (2 – w) (2 – w2) (2 – w10) (2 – w11) = 49 (b) ( x – y) (xw – y) ) (xw2 – y) = x3 – y35. If x = a + b , y = aw + bw2 and z = aw2 + bw , then prove that (a) x2 + y2 + z2 = 6 ab (b) x y z = a3 + b3MATHEMATICS 57
  • 20. Quadratic EquationsMODULE - I Algebra ANSWERS CHECK YOUR PROGRESS 2.1 Notes −4 1. (i) −2 3 , (ii) a − b, a + b 3 ab c (iii) − , (iv) 3 2, 2 c ab CHECK YOUR PROGRESS 2.2 3 ± 15 i 1± i 1. (i) (ii) 4 2 5 ± 43 i − 2 ± 58 i (iii) (iv) 8 6 1 2. –1, 2 CHECK YOUR PROGRESS 2.3 b 2 − 2ac (b 2 − 2ac)2 − 2a 2 c 2 1. (i) (ii) c2 c4 2. (i) 25x2 – 6x + 9 = 0 (ii) 625x2 – 90x + 81 = 0 4. q2 – 5p2 = 0 CHECK YOUR PROGRESS 2.4 1. (i) 3, 3w, 3w2 (ii) − 3, − 3w , − 3w2 (iii) 4, 4w, 4w2 (iv) − 4, − 4w , − 4w2 TERMINAL EXERCISE 3. (i) 5x2 – 8x + 5 = 0 (ii) 10x2 – 42x + 49 = 0 (iii) 25x2 – 116x + 64 = 0 58 MATHEMATICS