Hate using Quadratic Formula ? This could help!

[zabeer]

Like it or not, we have to solve quadratic equations in our exams - mostly as part of another question, rather than a direct "find the roots" question. Unless we could factorize in one look, most of us go for the traditional quadratic formula which could take time, with its division and square root steps.

Recently I came across another method - which was surprisingly simple, and was hidden in plain sight (at-least for me!)

This method is pioneered by Prof. Po-Shen Loh (CMU)

Here it goes.

For the equation, ax^2 + bx + c = 0, we know sum of the roots = -b/a and product of roots = c/a.

Let's take it one step ahead. consider m is the mid-point of the line connecting roots, we know it's (-b/a)/2.

Let's explain further with a simple example

Solve the roots for x^2 - 10x + 24 = 0

I know, you have factorized it in your head, and know the roots already! But let's keep it aside, and try the new one.

sum = 10
mid point between the roots = 10/2 = 5
which means the roots are at equal distance (say d) from 5 to either side.

(Root1 = 5 - d) ----- d ------- m = 5 ----- d ------- (Root2 = 5 + d)

so roots are (5 -d) and (5 + d)

we know product of roots = 24,
so (5 - d)(5 + d) = 24
25 - d^2 = 24
d^2 = 1

d = 1 (as it is a distance, we take positive value)
so roots are 5 -1 = 4 and 5 + 1 = 6

next one.

Solve the roots for 6x^2 - 19x + 10 = 0

Maybe bit tricky to factorize for a poor mortal, and time consuming with the traditional formula.

Will try with the above method

sum of roots = 19/6
mid point = 19/12
roots are at same distance (say d) from 19/12
so roots are (19/12 - d) and (19/12 + d).

Product = 10/6
(19/12 - d) * (19/12 + d) = 10/6
(19/12)^2 - d^2 = 10/6
d^2 = (19/12)^2 - 10/6 = (19^2 - 10 * 12 * 2)/144 = 121/144 = 11/12 (positive value only, as it's a distance)

roots = (19/12 - 11/12) = 8/12 = 2/3 & (19/12 + 11/12) = 30/12 = 5/2

one more ?

solve the roots for 4x^2 + 3x - 1 = 0

we know the drill :)

sum = -3/4
mid point = -3/8
roots are equidistant (say d) from -3/8
(-3/8 - d) and (-3/8 + d)

product = -1/4
(-3/8 - d) * (-3/8 + d) = -1/4
(-3/8)^2 - d^2 = -1/4
d^2 = 9/64 + 1/4 = (9 + 16)/64 = 25/64
d = 5/8 (take only positive value, as it's a distance)

roots = (-3/8 - 5/8) = -1, & (-3/8 + 5/8) = 1/4

what if roots are imaginary?

solve the roots for x^2 + x + 1 = 0

sum = -1
mid point = -1/2
roots are equidistant (say d) from -1/2
roots = (-1/2 - d) and (-1/2 + d)

product = 1
(-1/2)^2 - d^2 = 1
d^2 = 1/4 - 1 = -3/4
d = i√3/2
roots are -1/2 ± i√3/2

Can you solve the roots for x^2 - 2x + 8/9 = 0 (maybe, even without a pen! :) )

Happy Learning!