Lecture 1.3 methods of solutions of quadratic equations
- 1. LECTURE NO 3, METHODS OF SOLVING QUADRATIC EQUATIONS Solutions of the general quadratic equation 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0………………………….(A) 1. ‘Difficulty-symbol-solution’ Method We have to arrive at one solution of the general quadratic equation 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0. The roots to be found are 𝛼 = − 𝑏 2𝑎 ± √𝑏2−4𝑎𝑐 2𝑎 = − 𝑏 2𝑎 ± √𝐷 2𝑎 …………………..…(B) The method we describe is ‘Difficulty-symbol-solution’. The ‘difficulty’ was pointed out by someone ,say, to be the presence of linear term bx. We can not get rid of ‘bx’ term as such. We can replace x by another variable and get rid of linear term of that variable. We put a symbol 𝑥 = 𝑦 + ℎ and got an equation 𝑎(𝑦 + ℎ)2 + 𝑏(𝑦 + ℎ) + 𝑐 = 0………………………..……….(C) Or, 𝑎𝑦2 + 𝑦(2𝑎ℎ + 𝑏) + ℎ2 + 𝑏ℎ + 𝑐 = 0 If we put coefficient of y equal to 0, as we have to get rid of linear term in ‘y’, then 2𝑎ℎ + 𝑏 = 0 , or, ℎ = − 𝑏 2𝑎 . Further substituting this in (C) got the value of
- 2. 𝑦 = ± √𝑏2−4𝑎𝑐 2𝑎 . So 𝛼 = −𝑏+√𝑏2−4𝑎𝑐 2𝑎 , 𝛽 = −𝑏−√𝑏2−4𝑎𝑐 2𝑎 as square root has got two values, one + and the other . Finally it may be noted that we got , a. sum of the roots 𝛼 + 𝛽 = − 𝑏 𝑎 ……………………(D) and b. product of the roots 𝛼𝛽 = ( −𝑏+√𝑏2−4𝑎𝑐 2𝑎 ) ( −𝑏−√𝑏2−4𝑎𝑐 2𝑎 ) = 𝑐 𝑎 … … … ….(D’) We shall return to various aspects of this method afterwards while discussing quadratic functions. 2. Completing the square method Some student tried the equation (A) like this→ 𝑎𝑥2 + 𝑏𝑥 = −𝑐,→ 𝑥2 + 𝑏 𝑎 𝑥 = − 𝑐 𝑎 → 𝑥2 + 2𝑥 𝑏 2𝑎 = − 𝑐 𝑎 . The student wanted to take the linear ‘x term’ inside a ‘whole square and push all others to the right hand side. But the left side is not a complete square; this is the difficulty faced. To make it a complete square we have to add ( 𝑏 2𝑎 ) 2 to the left side ; obviously we have to add the same thing to the right hand side too, to maintain the equality sign. So we get
- 3. 𝑥2 + 2𝑥 𝑏 2𝑎 + ( 𝑏 2𝑎 ) 2 = 𝑏2 4𝑎2 − 𝑐 𝑎 Or, (𝑥 + 𝑏 2𝑎 ) 2 = ( √𝑏2−4𝑎𝑐 2𝑎 ) 2 , Or, 𝑥 = −𝑏±√𝑏2−4𝑎𝑐 2𝑎 = −𝑏±√𝐷 2𝑎 ………………………(D) Which is called the “quadratic formula” and 𝐷 = 𝑏2 − 4𝑎𝑐 is called discriminant’ of the quadratic equation. This discriminant shall be called for again and again for future use. Note that this method is also finding the difficulty and the solution is just by the side of the difficulty. Hundreds of years ago, an improvement of the method was given by Sridhar Acharya. Instead of dividing throughout by ‘a’, he multiplied throughout by 4a. This gives 4𝑎2 𝑥2 + 4𝑎𝑏𝑥 + 4𝑎𝑐 = 0 Or, 4𝑎2 𝑥2 + 4𝑎𝑏𝑥 + 𝑏2 = 𝑏2 − 4𝑎𝑐,adding 𝑏2 to both sides. This gives,(2𝑎𝑥 + 𝑏)2 = 𝑏2 − 4𝑎𝑐 . The advantage of this intelligent method is that the whole process is devoid of fractions until the last point. 3. A solution from, sum, product of roots. The sum and product of the roots ( vide D and D’) are
- 4. 𝛼 + 𝛽 = − 𝑏 𝑎 and 𝛼𝛽 = 𝑐 𝑎 respectively. So the problem of quadratic equation becomes essentially, the problem how to find two numbers whose sum and product are given. It would be quite easy if we find the difference of the roots, 𝛼 − 𝛽. From a well known formula in earlier classes, we have, (𝛼 − 𝛽)2 = (𝛼 + 𝛽)2 − 4𝛼𝛽 = (− 𝑏 𝑎 ) 2 − 4 𝑐 𝑎 = 𝑏2 − 4𝑎𝑐 𝑎2 𝛼 − 𝛽 = ± √𝑏2 − 4𝑎𝑐 𝑎 So, 𝛼 = 𝛼 + 𝛽 + (𝛼 − 𝛽) 2 = −𝑏 + √𝑏2 − 4𝑎𝑐 2𝑎 , 𝛽 = 𝛼+𝛽−(𝛼−𝛽) 2 = −𝑏−√𝑏2−4𝑎𝑐 2𝑎 . Taking ‘-‘ sign before the discriminant would produce the same roots in reverse order. The same solution we already got as before. 4. A solution from average of roots. From the sum formula, average of the roots ,say, ℎ = − 𝑏 2𝑎 = 𝛼 + 𝛽 2
- 5. Had the two roots been equal, each of them had been equal to the average. That is the meaning of average. In this case, one root shall be as much more than the average as the other root is less, average means just that also. So let the bigger root be 𝛼 = ℎ + 𝑘, and smaller root be 𝛽 = ℎ − 𝑘, where 𝑘 must be easily found out from the other relation 𝛼𝛽 = 𝑐 𝑎 . So we have, 𝑐 𝑎 = 𝛼𝛽 = (ℎ + 𝑘)(ℎ − 𝑘) = ℎ2 − 𝑘2 = 𝑏2 4𝑎2 − 𝑘2 So 𝑘2 = 𝑏2 4𝑎2 − 𝑐 𝑎 = 𝑏2−4𝑎𝑐 4𝑎2 , or, 𝑘 = ± √𝑏2 −4𝑎𝑐 2𝑎 Then the roots are 𝛼 = ℎ + 𝑘 = −𝑏+√𝑏2−4𝑎𝑐 2𝑎 and 𝛽 = ℎ − 𝑘 = −𝑏−√𝑏2−4𝑎𝑐 2𝑎 Simple solution of simple problem. There are many other methods and the list is not exhaustive. Then why should anyone learn many methods? The reason is, that only one method of solution may be learnt; but the other methods turn into just problems when differently worded. An improvement:
- 6. First reduce the equation to the form 𝑥2 + 2𝐵𝑥 + 𝐶 = 0 ( just divide 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0 by a; call 𝑏 𝑎 = 2𝐵, 𝑐 𝑎 = 𝐶.) Now 𝛼 = −𝐵 + √𝐵2 − 𝐶 and 𝛽 = −𝐵 − √𝐵2 − 𝐶.