As we know for a quadratic equation ax² + bx + c = 0 , where a is not = 0 , have many ways of solving them ( finding roots ), those methods include factorization (mid term splitting ) , completing squares and quadratic formula , now here is a method made by me for finding roots.
Take any quadratic Equation , for example
x² - x - 6 = 0
here a = 1 , b = -1 , x = -6
α+ß = -b/a = 1
αß = c/a = -6
now
(α + ß )² = α² + ß² + 2αß
1 = α² + ß² - 12
α² + ß² = 13
(α - ß)² = α² + ß² - 2αß
(α - ß)² = 13 + 12
α - ß = ±5
on subtracting the equations
Case 1 where α - ß = 5
(α + ß) - (α - ß) = 2ß = -4 , ß = -2 , so one root of the polynomial is -2 , α - 2 = 1 , α = 3 , this is another root .
Case2 where α - ß = -5
(α + ß) - (α - ß) = 2ß = 6 , ß = 3 , so one root of the polynomial is 3 , α + 3 = 1 , α = -2 , this is another root . so if we take any one case , we will get the roots of the polynomial
Take any quadratic Equation , for example
x² - x - 6 = 0
here a = 1 , b = -1 , x = -6
α+ß = -b/a = 1
αß = c/a = -6
now
(α + ß )² = α² + ß² + 2αß
1 = α² + ß² - 12
α² + ß² = 13
(α - ß)² = α² + ß² - 2αß
(α - ß)² = 13 + 12
α - ß = ±5
on subtracting the equations
Case 1 where α - ß = 5
(α + ß) - (α - ß) = 2ß = -4 , ß = -2 , so one root of the polynomial is -2 , α - 2 = 1 , α = 3 , this is another root .
Case2 where α - ß = -5
(α + ß) - (α - ß) = 2ß = 6 , ß = 3 , so one root of the polynomial is 3 , α + 3 = 1 , α = -2 , this is another root . so if we take any one case , we will get the roots of the polynomial
Why not to use FACTORIZATION ???
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