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Model Lecture

Published in: Mathematics
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  • Hemant P

    • Mumbai
    • 15 Years of Experience
    • Qualification: M.Sc
    • Teaches: Mathematics, IIT JEE Mains, IIT JEE Advanced
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This is model lecture on Quadratic equations.

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    Model lecture - Theory of equations 1. Introduction: Difference between equation, expression and identity. Expression: An expression is a relation between two variables which yields infinite values for the two variables. An example will clarify this; for example y — x is a relation between y coordinate and x coordinate i.e. a general relation between two or more variables is called an expression. y =3x —x + 1 is an example of quadratic equation. Equation: Relation between variables/variable with a sign of equality fetching a particular value for the variable is called an equation. For example 2x+3 = 0 is a equation of degree one and the particular value is x 1 2/3 .Similarly 2x +3x+1=0 is an quadratic equation, with particular values = x 1 and x = 2 Identity: An identity is an expression true for all values of various variables. For example (a + = a2 + b2 + 2ab is an identity which is true for all values of x. In the context of quadratic equations if a, ß are the roots of the quadratic equation ax +bx+ c = Othen ax2 +bx+c is a quadratic identity. We take a particular example to explain all the concepts. y = x —5x+6 is an expression, y = x —5x+6 = 0 is an equation which yields the particular values x = 2 and x = 3 ,while (x 3) x 2 5x+6 is an identity which is true for all values of x. We will find application of above property in the following problems. Example 1. Let the complex roots of the equation 1337* 1336x4 +1335* 1334* +1333* 1332 a, b, c, d, e .Find the value of (1 + a)(l +b)(l + c)(l +d)(l + e) ? Solution: Since a, b, d, e, fare the roots of the equation we get 1337x5 -1336x4 +1335x3 -1334x2 vx Now putting x = i = NCI we get 5 3 + 1335 1337 Or (1337 + 1336 + 1335 + 1334 + 1333 + 1332) 5 1337 (1 + a) (I + b) (1 + c) + d) -k e 5 5 Or 5 1337 (1 + -k b) (1 -k c) (1 -k -k e -(1337+1332) — 2669 2 2 5 5 2669 (1337 + 1332) 2 2 1337 e)
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    We take few more examples Example 2. If a, ß, y, are the roots of the following cubic equation 4x3 + 5x2 + 7 x + 9 Find the value of (1 + a2 +ß2 + ) Solution: Since a, ß, y, are the roots of the equation 4x3 +5x2 +7x+9 Y x Putting x = i = we get Or —4i —5 +7i+9 4 (i Y) Taking modulus we get 13i +41 14(i a) Or 32 +42 = 4 25 16 Example 3. If a, ß, are the roots of the equation (x —a)(x —b) = c , Then a, b, are the roots of ...? Solution: Since a, ß, are the roots of the equation (x —b) =c, a)(x p) (x a)(x b) c —(x a)(x b) (x a)(x ß)+C :.a, b, are the roots of (x a)(x ß)+c=0 2. Sridharacharya's Formula: Solution of quadratic equation using method of completing squares A quadratic equation is an equation in second degree of the form ax + bx+ c where a # 0, and a, b, c, e R .We wish to find the solution of the quadratic equation ax + bx+ c .We will do the same by using Shridharacharya's formula or method of completing squares. Multiplying both sides by 4a we get 4a2X2 -k 4abX -k — O Comparing with x 2 + y +2xy by completing the squares we get + b) + b2 —4ac A; where A = Discriminant; i.e. quantity which discriminates between nature of the root s of the quadratic equation.
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    —b± b2 —4ac Or x = ...(1) which is the solution of the quadratic equation. Now we can easily see that above nature of roots of the above equation is dependent on the term in the square root A A b2 - 4ac — Discriminant b2 — 4ac > O Roots are disänct& real b2 — 4ac < O Roots are Non-real complex or imaginary b2 — 4ac = O Roots are real and equal b2 — 4ac = Perfect square Roots are disånct and rational We now see some examples of the above. Example 4. If a, b, c, are the sides of the triangle and x2 —2(a + b + c) x + 3(ab + bc + ca)l = O ; xe R then the range of X will be? [11T JEE 20061 Solution: Roots are real Discriminant = A b2 —4ac > 0 {2(a + b + c)) 2 4• 1 • 3(ab + bc + 2 2 2 2 Now a, b, c are sides of a triangle cosA 2bc Similarly c2 + a2 —b2 < 2ac And a2 + b2 — c2 < 2ab a2 < 2bc Alternately we can square and add the identity a c < b and its variants (Difference of the two sides is less than the third side). 2 Adding we get a2 + b2 + c 2 < 2(ab + bc + ca) 2 Or 31-2
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    (5+2U6)X2 ? Example 5. Solve for 'x' Solution: Let (5 + 3 Y 25-24 10 1 . The equation becomes y + 10+ 100-4 5 + 2A./ö [Using (1)] 2 [Using (2)] We can also solve the same problem by power of patterns or one step method. They are not short cuts but only application of symmetry and patterns .We can solve many problems through these smart methods only by keeping symmetry of the problems in mind. Rewriting RHS as 10 (5+ 2N6) + (5 — 2$) and comparing indices we get x2 3=+1 > x 2 —3+1—4 or 2 2, or x +2 (4+un-s)X ? Example 5. Solve for 'x' Solution: Let us try to use what we have learned just now .This does not look straight forward. Well it is. We often tell to our students that observation is the key to problem solving. A little mental math will tell that (4+uS)2 Since radicals will cancel out and 31>
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    10b Iyl z z(Z+2) Or z2 +2z—3=0 Or z2 +3z —z —3=0 +3) I(z +3) —0 Or z(z +3) I(z+3) Or (z +3)(z 1) —0 10b I yl 1 Or 10b I yl 1 8 Since 4x 2 3 3 or z 3 1 or 4x +1=2 or 4x We now use the power of patterns to solve the problem in one step. We try to explain the essence of the method .Mathematics is a symmetric science i.e. RHS is exactly the same as LHS for some value of 'x'. —On RHS we must have product of two logarithms similar to that on LHS. Rewriting RHS as 10b 8 • 10b 2 and the problem is solved! 10b Or on comparing we get 4x +1=2 or 4x = 1 3. Vieta's Theorem: Relationship between coefficients and roots of a quadratic equation Let a, ß, be the roots of the of the quadratic equation ax +bx+ —a(x a)(x p) ax 2 + bx+c Or a(x2 (a + p) + •aß) ax 2 + bx+ c 2 a By comparison we get b and aß a a We can also demonstrate the above relationship by adding and multiplying the two values of 'x' obtained in (1). We now take few more examples to get used to of the above results. Example 7. If roots of —10ax—11b=O are c, d and roots of x 2 find the value of a+b+c+d ? Solution: are roots of x 2 -10ax -11b 0 lid - — O are a, b then [11T JEE 20061
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    . Using Vieta's theorem we have sum of roots And product of roots = cxd=11b Similarly a, b are roots of x 2 -10cx - 1 Id 0 Sum of roots = a+b =10c And products of roots Id Now (1)+(3 ) gives a-kb-kC+d= 10 (a-PC) Also (1)—(3) gives Or (d—b)=ll(a—c) Since 'c' and 'a' are roots of (A) and (B) we get Putting 'c' in (A) we get c2 -10ac -11b —0 = C-kd (4) =10a ...(1) ...(B) ...(2) ...(3) ...(5) ...(6) Also putting 'a' in (B) we get a2 -10ac-1 Id :. (5)—(6) gives c2 —a2 11 (b — d) Or (c—a)(c+a) Il(b d) (c—a)(c+a) llxll(c—a) c + a — 121 Putting above value in (3) we get a-kb-kC+d= I ox 121 = 1210 From We will learn another method to find values of certain symmetric equations with ease. Before that, let us know something about symmetric functions. This method is also used in transformation of equations. In ordinary language symmetric functions are functions in two or more variables which remain unchanged when we replace the variables in cyclic order. For examplea+ß,aß,a +ß3 are symmetric functions. If sign of the expression is changed then the function is called anti symmetric function. 3 1 Example 8. If a, ß are roots of x 2 + 4x +1 = O find the value of 3 1 Solution: We will try to find try to find out the solution by usual method and see that it is difficult to find the solution. By Vieta's theorem ct+ß = —4 and 3 1 3 1 1 3 1 ...(1) 1 1 1 1 3 ...(2)
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    Now we find the values Now and 4+2 Also (a + I) • + I) a • + a + + I —4 -k I -k I = —2 Also :. Putting above values of (3) and ...(4)in (2) we get ....(3) 3 2 Also a 4 + ß4 — (a2 + 2a2ß2 {(a + 5 2 2a.ß}2 2a2ß2 {(-4)2 2-1 We will try to find the same values by method of transformation. To findthe values of we try to find a equation whose roots are of the form Another transformation of the form z will give the required equation. . The transformation x — Or (I-y)2 —0 I-y gives I-y +1=0 or 2 y 2 — 2 y — 1 = 0 We now perform the second transformation z = y > y 2z3 —2z3 —1—0
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    2 Or 2z3 1 2z3 —1 Cubing both sides gives 2 Or 8z2 —8z —3•2z3 Or 8z2 —8z —12z •1 1 • 2z3 1 2 1 2z3 3 1 1 Using (5)) 3 1 and 0 has 1 The transformed equation 8z —20z as its roots. 3 1 3 1 5 20 8 2 You can use any of the methods depending of personal preferences. Also one may change order of 3 1 transformation i.e. you may use x = y first and y —.The equivalent transformation x z is important ,the intermediate steps can be chosen on convenience. 1 Similarly for a 4 + ß 4 4 we must use the transformation y —x i.e. x = y 2 1 1 1 Or y 4 1 Squaring both sides we get 1 y + 16 -k • Y 4 | ...(6) 1 or Y+16C+2V.( 1) 1 [Using4y4 Or y +14.11 1 1 Squaring again we get 196y + y 2 + 2 y Y 2 194 y -k I O 1 from (6) ] 4 This is the required transformed equation with roots a and (-194) . As before Sum of roots = a 4 +ß 4 194 1

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