Quadratic equations - standard form, formula for solving

1. Quadratic Equations Algebra In one unknown(variable)

3. 3x2+11=0

6. 4x2+5x=0 (There are 2 terms on LHS)

7. 2x2-50=0 (There are 2 terms on LHS)

9. Quadratic equations Solving quadratic eqns. Solving a simple quadratic equation-1 Quadratic equation without x term: Ex: x2=4 ⇒x2-4 = 0 (All terms transferred to LHS. There is no x term) ⇒x2-(2)2 = 0 (This is in the form a2-b2) ⇒(x+2).(x-2) = 0 [∵ a2-b2 =(a+b).(a-b)] ⇒ (x+2) =0 OR (x-2) = 0 (By zero product rule - if product of two factors is zero, then at least one factor is zero) ⇒ x = -2 OR x = 2 Hence there are two solutions for this quadratic equation. In general for every quadratic equation there will be two solutions. Yes! The equation is now reduced to a product of two linear expressions. Hence each can be easily solved separately. Click to Continue

10. Quadratic equations Solving quadratic eqns. Solving a simple quadratic equation-2 Quadratic equation without constant term: Ex: 4x2+5x=0 ⇒x(4x+5) = 0 (Taking out the common factor x from eqn ) ⇒ x =0 OR (4x+5) = 0 (By zero product rule - if product of two factors is zero, then at least one factor is zero) ⇒ x = 0 OR x = -5/4 Hence there are two solutions for this quadratic equation. In general for every quadratic equation there will be two solutions. Yes! The equation is now reduced to a product of two linear expressions. Hence each can be easily solved separately. Click to continue [Teaser: Do you think there will be, in general, 3 solutions to a cubic equation –Yes! You are right, there will be. And by extension there will be n solutions for a n-the degree equation]

11. Quadratic equations Solving quadratic eqns. Points on quadratic equations The standard form of a quadratic equation is ax2+bx+c = 0, where x is unknown (variable) and a, b, and c are all real number constants and a ≠ 0. [a, b, c are called coefficients of the quadratic equation] A quadratic expression can be factorized into two linear expressions – as ax2+bx+c = (px+q)(rx+s), and hence any given quadratic equation can be changed to the form (px+q)(rx+s) =0. [How to factorize – we will see later] Now Zero product rule is applied to solve the equation. For example if (x+3)(x-2) =0, then (x+3)=0 or (x-2)=0; which means x=-3 or x=2. Every quadratic equation has two solutions, which are also called the roots of the equation. If a =0, then the equation is reduced to bx+c=0, which is no longer quadratic equation, but simply a linear equation. Click to continue

13. By FormulaFactorization method is suitable for simple equations, where the coefficients in the equation are simple and easy to factorize. Formulas method is suitable for all types of problems, and can be applied systematically to solve any equation, and directly gives solutions.

14. Quadratic equations Solving by Factorization Solving quadratic equations by factorization This is a trail and error method (that means it is not systematic). It is suitable when the coefficients a, b and c are small & simple, and can be easily factorized by hand. The factorization is done as follows: Factorize a.c into p and q such that b = p+q Then, the quadratic expression can be easily factorized. Examples will make the concept clear

16. Transpose all the terms to LHS to get the equation into standard form

17. Factorize the expression on the LHS

19. Quadratic equations Solving by Factorization Worked out examples-1 Problem: 2x2-7x=39 Solution: (continued) 2x2-7x-39=0 ⇒ 2x2 +6x-13x-39=0 (bx term is split into px+qx) ⇒ 2x(x+3)-13(x+3)=0 ⇒ (x+3)(2x-13) [by taking out common factor (x+3)] ⇒ (x+3)=0 OR (2x-13) = 0 [By zero product rule] ⇒ x = -3 OR 2x =13 ⇒ x = -3 ORx =13/2 [These are solutions of given quadratic equation] [Tip:-It is always a good idea to check the correctness of solutions by substituting the answers back into original equation, and see if it is satisfied] Taking out common factor 2x from the first two terms. Click to continue Taking out common factor -13 from the last two terms Click to continue Yes! The equation is now reduced to a product of two linear expressions. Hence each can be easily solved separately. Click to continue

20. Quadratic equations Solving by Factorization Worked out examples-2 Problem: Solve Analysis: Before solving let us analyze the problem. The given problem is not a quadratic equation (at least in the given form). There are fractions here. Hence we try to eliminate fractions first. How do we do it? Take GCD for LHS fractions and make it a single fraction. On RHS convert the given mixed number 2½ into improper fraction of 5/2. Then cross multiply to eliminate fractions.

21. Quadratic equations Solving by Factorization Worked out examples-2 Problem: Solve Solution: By cross multiplication. Click to continue By combining fractions on LHS – first take GCD of (x-1) and x for the resultant fraction. Click to continue

22. Quadratic equations Solving by Factorization Worked out examples-2 Solution: (Continued) ⇒ 4x2-4x+2 = 5x2-5x (Simplifying both sides) ⇒ -x2+x+2 =0 (By transferring all terms to LHS) ⇒ x2-x-2 =0 (By multiplying both sides by -1, so we get positive x2 term) ⇒ x2-2x+ x -2 =0 ⇒ x(x-2) + (x-2)=0 (By taking x common from first two terms) ⇒ (x+1)(x-2) = 0 (By taking x-2 common ) ⇒ (x+1) = 0 OR (x-2) =0 (By zero product rule) ⇒ x=-1 OR x=2 Now factorize axc =(1)x(-2)=-2, into two factors p and q such that p+q = b (i.e. -1). Factors of -2 are : -1 and 2 (and -1+2 =1 ≠ -1 (wrong) 1 and -2 (and 1-2 = -1 =-1 (b, hence OK) Click to continue Now this is in standard form of quadratic equation. By comparison the coefficients are a=1; b=-1; c=-2. Click to continue

23. Quadratic equations Solving by Factorization Worked out examples-3 Problem: Find the quadratic equation whose solution set is (-2,3) Analysis: It is a reverse(inverted) problem. Here we are given the pair of solutions (as a quadratic equation has two solutions), and we have to find the original quadratic equation. Here is a hint. See the solution procedure for a normal problem (i.e., quadratic equation is given, and you have to find the solution set). You can simply start at the bottom of that procedure and move up the steps. Consider the solution steps for the problem 2x2-7x-39=0, which you have already solved. I have given the steps, which are already done by you in the next slide. Now if the solution set is given the steps can be numbered from bottom to top as shown

24. Quadratic equations Solving by Factorization Worked out examples-3 Analysis of the Solution of previous problem: 2x2-7x-39=0 -Required equation 2x2 +6x-13x-39=0 -Step 6 2x(x+3)-13(x+3)=0 -Step 5 (Expanding) (x+3)(2x-13)=0 -Step 4 (x+3)=0 OR (2x-13) = 0 -Step 3 x = -3 OR 2x =13 -Step 2 x = -3 ORx =13/2 -step 1[These are solutions of given quadratic equation] Now you can see that for a inverted(reversed) problem, the solution steps can also be simply reversed. That is great, and this is much simpler than original (normal) problem. No great mathematics involved here. Have you noticed the reversed implies symbol Click to continue

25. Quadratic equations Solving by Factorization Worked out examples-3 Back to the Problem: Find the quadratic equation whose solution set is (-2,3) Solution: Solution set =(-2,3) [Given] ⇒ x=-2 OR x=3 ⇒ x+2=0 OR x-3=0 [Two linear equations] ⇒ (x+2).( x-3)=0 [Now we take the product of the two linear equations, where as in the normal problem we factorize into two linear equations] ⇒ x2 +2x-3x-6=0 [Expanding the equation] ⇒ x2 -x -6=0 [Simplifying LHS] – required equation

26. Quadratic equations Solving by Factorization Worked out examples-4 Problem: Find the quadratic equation whose solution set is (α,β) Analysis: It is same as previous problem, but we are trying to find a general solution Solution: Solution set = (α,β) [Given] ⇒ x= α OR x = β ⇒ x-α=0 OR x-β=0 [Two linear equations] ⇒ (x-α).( x-β)=0 [By taking the product of two linear equations] ⇒ x2 -αx- βx+αβ =0 [Expanding the equation] ⇒ x2 –(α+ β)x+αβ=0 [Simplifying LHS] – required equation [It simply means that for given roots α and β, the quadratic equation is defined by the coefficients a=1, b= -(α+ β), and c = αβ]

27. Quadratic equations Solving by Factorization Worked out examples-4 Some examples of the above type problems What is the quadratic equation whose solution set is (3,4) x2 –(3+4)x+3.4=0 [substituting α = 3, β=4] Ξ x2 –7x+12=0 [On simplification] 2) What is the quadratic equation whose solution set is (-5,7) x2 –( - 5+7)x+(-5).(7)=0 [substituting α = -5, β=7] Ξ x2 –2x-35=0 [On simplification] Note: Be very careful while substituting α and β values and associated signs. In a descriptive problem do not write the answer by direct substitution of α and β values, but show full derivation.

28. Quadratic equations Solving by Factorization Worked out examples-5 Problem: If x =2 and x=3 are roots of the equation 3x2 –2mx+2n=0; find the values of m and n. Analysis: This is another slightly twisted problem. Here roots of the quadratic equation given, but coefficients are not given. Hence what we do, we substitute two root values in the given equation, giving us two equations. Now we have two equations in two unknowns (m and n ), and we already know how to solve them . Now we proceed to solve the problem

29. Quadratic equations Solving by Factorization Worked out examples-5 Problem: If x =2 and x=3 are roots of the equation 3x2 –2mx+2n=0; find the values of m and n. Solution: Substituting the 1st root (i.e., 2) in the give equation we get 3(2)2 – 2m(2) +2n=0 ⇒ 12 – 4m + 2n =0 ⇒ 4m –2n =12 [By rearrangement] Now substituting 2nd root (i.e., 3) in the given equation we get 3(3)2 – 2m(3) +2n=0 ⇒ 27 – 6m + 2n =0 ⇒ 6m –2n =27 [By rearrangement] Now we have two equations (linear) in two unknowns - give us 2m=15 ⇒ m =15/2 Substituting this value of m in eqn we get 4(15/2) -2n =12 ⇒ 2n = 18 ⇒ n=9; Hence m=15/2 and n=9 1 2 2 1 1

30. Quadratic equations Solving by Formula Solving quadratic equations by Formula This is the general method that can be applied routinely for any quadratic equation. The roots of the quadratic equation ax2+bx+c = 0, where a ≠0 can be obtained by using the formula:

31. Quadratic equations Solving by Formula Solving quadratic equations by Formula Proof: ax2+bx+c = 0 [Given] ⇒4a(ax2+bx+c ) =4a(0) [Multiplying the entire eqn by 4a] ⇒ 4a2x2+4abx+4ac = 0 ⇒ (2ax)2+2(2ax)(b)+4ac =0 [4a2x2 is written as (2ax)2 and 4abx is written as 2(2ax)(b)] ⇒ (2ax)2+2(2ax)(b)+b2-b2+4ac =0 [b2 is added and subtracted to LHS] ⇒ (2ax+b)2 -b2+4ac =0 ⇒ (2ax+b)2 =b2-4ac ⇒ (2ax+b) = ± ⇒ This part of the expression is in the form of p2+2pq+q2,where 2ax =p and b = q, and hence this part can be written as (p+q)2 Click to continue

32. Quadratic equations Solving by Formula Worked out examples-1 Problem: Solve the quadratic equation using formula: 5x2-2x-3= 0 Solution: Comparing the given equation with standard form, ax2+bx+c = 0 we get a = 5, b = -2, c = -3 Substituting these values in the formula [Carefully substitute values in the formula, observing signs duly. Do not attempt short cuts when substituting values in the formula] ⇒x= 1 OR -3/5 It is wonderful that this number within the square root is perfect square. But this does not always happen Hence, when b2-4ac >0, we have two distinct real roots

33. Quadratic equations Solving by Formula Worked out examples-2 Problem: Solve the quadratic equation using formula: x2+6x+9= 0 Solution: Comparing the given equation with standard form, ax2+bx+c = 0, we get a = 1, b = 6, c = 9, Substituting these values in the formula ⇒x= -3 Now the value b2-4ac is zero Hence when the value b2-4ac is zero, there is only one root

34. Quadratic equations Solving by Formula Worked out examples-3 Problem: Solve the quadratic equation using formula: 5x2-6x+7= 0 Solution: Comparing the given equation with standard form, ax2+bx+c = 0, we get a = 5, b = -6, c = 7 Substituting these values in the formula ⇒x has no real roots Now the value b2-4ac is negative, and there is no square root of a negative number Hence when the value b2-4ac is negative ,the quadratic equation has no solutions in the set of real numbers.

35. Quadratic equations Solving by Formula Worked out examples-4 Problem: Solve the quadratic equation using formula: 2x2+6x+3= 0 Solution: Comparing the given equation with standard form, ax2+bx+c = 0, we get a = 2, b = 6, c = 3 Substituting these values in the formula Now the value b2-4ac is positive, but it is not a perfect square Hence when the value b2-4ac is positive, the quadratic equation has two real roots. But since b2-4ac is not a perfect square, the roots are irrational

38. Quadratic equations Nature of roots You need not remember above results. You can simply and easily derive the above from the formula for equation of roots Case I: B2-4ac >0 Sub case I(i) – b2-4ac is also a perfect square, then the highlighted expression also evaluates to an integer and hence the roots also become rational. Click to continue Sub case I(ii) – b2-4ac is not a perfect square, then the highlighted expression evaluates to an irrational number and hence the roots also become irrational. Click to continue Case II: B2-4ac =0 Case III: B2-4ac <0 Then this expression evaluates to some positive value, say p. Then the roots are (-b+p)/2a and (-b-p)/2a – and hence two real roots. Then the highlighted expression value, say p, becomes 0. Hence whether the roots (-b+0)/2 or (-b-0)/2 are same. Hence only one root Then the highlighted expression value, say p, can not be evaluated, as there are no real suare roots for a negative number.

39. Quadratic equations Nature of roots Worked out examples-5 Problem: Without solving the equation 7x2-9x+2= 0, comment on the nature of roots Solution: Comparing the given equation with standard form, ax2+bx+c = 0, we get a = 7, b = -9, c = 2 Substituting these values in the formula: Discriminant D = b2-4ac D = (-9)2 – 4(7)(2) ⇒ D = 81 – 56 ⇒ D = 25 Since D >0; The roots are real and unequal (There are two distinct real roots) Also D is perfect square as 25 = 52, Hence roots are also rational. Hence the roots are rational, real, and unequal.

40. Quadratic equations Nature of roots Worked out examples-6 Problem: Without solving the equation 6x2-13x+4= 0, comment on the nature of roots Solution: Comparing the given equation with standard form, ax2+bx+c = 0, we get a = 6, b = -13, c = 4 Substituting these values in the formula: Discriminant D = b2-4ac D = (-13)2 – 4(6)(4) ⇒ D = 169 – 96 ⇒ D = 73 Since D >0; The roots are real and unequal (There are two distinct real roots) Also D is not a perfect square as 73 can not be written as square of an integer, Hence roots are also irrational. Hence the roots are irrational, real, and unequal.

41. Quadratic equations Nature of roots Worked out examples-7 Problem: Without solving the equation 25x2-10x+1= 0, comment on the nature of roots Solution: Comparing the given equation with standard form, ax2+bx+c = 0, we get a = 25, b = -10, c = 1 Substituting these values in the formula: Discriminant D = b2-4ac D = (-10)2 – 4(25)(1) ⇒ D = 100 – 100 ⇒ D = 0 Since D =0; The roots are real and equal (There is only one real root) Hence the roots are real and equal.

42. Quadratic equations Nature of roots Worked out examples-8 Problem: Without solving the equation 2x2+8x+9= 0, comment on the nature of roots Solution: Comparing the given equation with standard form, ax2+bx+c = 0, we get a = 2, b = 8, c = 9 Substituting these values in the formula: Discriminant D = b2-4ac D = (8)2 – 4(2)(9) ⇒ D = 64 – 72 ⇒ D = -8 Since D < 0; There are no real roots Hence there are no real roots(The roots are imaginary).

43. Quadratic equations Nature of roots Worked out examples-9 Problem: Find the value of ‘m’ , if the roots of the following quadratic equation are equal: (4+m) x2 +(m+1) x + 1 =0 Analysis: In this type of problems, we are given a quadratic equation in which one/more coefficients (a, b, c) of the given equation are not given, but expressed in some unknown term. For example in the above equation coefficients a and b are expressed in terms of ‘m’, which is an unknown. But we are given a condition about the nature of roots. For example in this problem we are given that the roots of the equation are equal. Hence we are given that b2-4ac =0. Now by substituting the coefficients in this condition we get another equation (some times linear, sometimes quadratic]. Now we solve this new equation for the unknowns in the coefficients.

44. Quadratic equations Nature of roots Worked out examples-9 Problem: Find the value of ‘m’ , if the roots of the following quadratic equation are equal: (4+m) x2 +(m+1) x + 1 =0 Solution: Comparing the given equation with standard form, ax2+bx+c = 0, we get a = (4+m), b = (m+1), c = 1 Roots of the quadratic equation are equal [Given] ⇒ b2-4ac =0 ⇒ (m+1)2 – 4(4+m)(1) = 0 [Substituting values of a,b,c in the eqn] ⇒ (m2+2m+1) – (16+4m) = 0 [Expanding terms] ⇒ m2 -2m – 15 =0 [By simplification] [Now this is a quadratic equation in variable ‘m’. Hence solve it by usual methods – by factorization or by formula] By factorization Now by Comparing the given equation with standard form, ax2+bx+c = 0, [contd…..]

45. Quadratic equations Nature of roots Worked out examples-9 Solution: [contd…..] we get a = 1, b = -2, c = -15 Now Factorize (a.c) into two factors p and q such that p+q = b a.c = (1).(-15) = -15 (- 1x3x5) [displaying the prime factors] Possible values of p and q are Hence given equation can be written as: m2 -5m + 3m– 15 =0 [Or you can also write as m2 + 3m -5m– 15 =0] ⇒ m(m-5)+3(m-5) = 0 [By taking m common from first two terms and 3 common from last two terms] ⇒ (m+3)(m-5) = 0 [By taking common the factor (m-5)] ⇒ m= -3 OR m=5 [ For both these values of ‘m’ the original equation has equal roots] It is a good idea to check the results, by actually substituting one/both values in the original equation , and see if indeed it has equal roots. Click to continue This table is shown only for your clarity. In actual problem solving , you need not show these computations, but you can workout these in margins Click to continue

46. Quadratic equations Nature of roots Worked out examples-10 Problem: Solve the following equation for x, and give your answer correct to 2 decimal digits: Analysis: First the given problem has fractions. Hence first eliminate fractions. Combine fractions on LHS by taking GCD and then cross multiply both sides of eqn, and simplify the given eqn into a standard form. Next we are asked to solve the equation correct to 2 decimal places. That is we may get roots that are irrational (i.e., √2 or √5 etc). You may look up square root values from the tables.

47. Quadratic equations Nature of roots Worked out examples-10 Solution: Comparing the given equation with standard form, ax2+bx+c = 0, we get a = 2, b = -7, c = -1 Substituting these values in the formula We get [contd….]

48. Quadratic equations Nature of roots Worked out examples-10 Solution: [contd….] 57 is not a perfect square, nor can any factors be taken out of the root sign (radical). Click to continue Since these are approximate solutions, you can not check these values by substituting them into original equation. The result will not be exactly zero. Click to continue

51. Quadratic equations Equations Reducible to quadratic equations Worked out examples-11 Problem: Solve 2x4-5x2+3 = 0 Solution: 2x4-5x2+3 = 0 [Given] ⇒ 2(x2)2-5(x2)+3 = 0 [Writing equation in terms of x2] ⇒ 2t2-5t+3 = 0 [By substituting x2 = t] ⇒ 2t2-3t-2t+3 = 0 [Splitting -5t into -3t and -2t for factorization] ⇒ t(2t-3) - (2t-3) =0 [Factorizing] ⇒ (t-1)(2t-3) =0 [Factorizing] ⇒ t = 1 OR t = 3/2 [Solutions for dummy variable] ⇒ x2 = 1 OR x2 = 3/2 [Replacing dummy variable with original one] ⇒ x2 – 1=0 OR x2 -3/2 = 0 [Two quadratic equations] ⇒ (x +1)(x-1) = 0 OR [x+√(3/2)][x-√(3/2)]=0 [∵ a2-b2 = (a+b)(a-b)] ⇒ x = 1 OR x = -1 AND x = √(3/2) OR x = -√(3/2)

52. Quadratic equations Equations Reducible to quadratic equations Worked out examples-12 [Advance problem] Problem: Solve (x2+3x)2 -(x2+3x)-6 = 0 Analysis: If you expand the given equation, you get 4th degree equation. But without expanding ,you can observe that the expression (x2+3x) can be substituted by a dummy variable t, to reduce the given equation into quadratic form. Substitute t for (x2+3x). The equation then reduces to: t2-t-6. Now this is a quadratic equation. On solving this equation Given eqn can be factorized as (t-3)(t+2) =0 we get t = 3 or t = -2; [two solutions for dummy variable] Now we solve (x2+3x)= 3 and (x2+3x)=-2 (two simple quadratic eqns) Solving each we get And x=-1 and -2 [The detailed steps of solution left to student as exercise]

53. Quadratic equations Equations involving radicals (square roots) [Advanced problems] Sometimes the given equation involves radicals (i.e., square roots etc). In such cases, the radicals must be eliminated by raising both sides of the equation to required power (exponent). For example if the given equation is: The above equation is rationalized (radicals eliminated) as follows: Always check the solutions of quadratic equations (obtained by rationalizing) in the original equation – as some solutions may not satisfy the original equation. They have to be discarded. Click to continue Now this is quadratic equation, which can be solved by methods already studied Click to continue

54. Quadratic equations Equations involving radicals Worked out examples-13 [Advance problem] Problem: Analysis: Since the equation contains radicals (square root terms) we can rationalize the equation by taking square on both sides and simplifying it to a quadratic equation. But squaring always generates large terms. If you observe that the LHS has two terms, which are reciprocals of each other. That is if one term is y, then other term is 1/y. This substitution makes the equation simpler to solve. Hence solve this equation by substitution.

55. Quadratic equations Equations involving radicals Worked out examples-13 [Advance problem] Solution: [By taking GCD on LHS and converting mixed number into improper fraction on RHS] ⇒6(y2+1) = 13y [On cross multiplication] ⇒ 6y2- 13y+6 = 0 [ on rearranging – we get a quadratic equation] ⇒ 6y2- 9y-4y+6 = 0 [-13y is split for factorization as -9x-4 = 36, which is product of 6x6] ⇒ 3y(2y-3) – 2(2y-3) = 0 ⇒ (3y-2)(2y-3) =0 ⇒ y = 2/3 OR y = 3/2 [Now we can replace dummy variable for original variable] [Contd……]

56. Quadratic equations Equations involving radicals Worked out examples-13 [Advance problem] Solution: [Contd….] ⇒ 9x = 4-4x OR 4x = 9-9x [On cross multiplication] ⇒ 13x = 4 OR ⇒ 13x = 9 [On simplification] ⇒ x = 4/13 OR x = 9/13 Now check the results in the original equation Substitute x = 4/13 [Contd….]

57. Quadratic equations Equations involving radicals Worked out examples-13 [Advance problem] Solution: [Contd….] Similarly you can check that other solution also satisfies the original equation

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