Quick/Simple Approach To Solving Quadratic Equation Involving Factorization Method


If; ax² + bx + c = 0------------(i)
ax² + (α+β)x + c = 0 ----(ii)

Note: From equation (ii) b= α+β
Product of the coefficient of x² 'a' and the constant 'c' = a.c
Considering a.c: the two factors of a.c which when added gives the coefficient of x
Let the two factors be α and β
The solution to the quadratic equation will now be :
x=(additive inverse of ∝)/(coefficient of x² i.e a)
OR
x=(additive inverse of β)/(coefficient of x² i.e a)

In short;

x=∝_(add.inv)/a
OR
x= β_(add.inv)/a
a is the coefficient of x² , ∝and β are the factors of the product of the coefficient of x² i.e a, and the constant 'c'
Examples
Consider these two sets of quadratic equation :
x² - 3x - 4 = 0 ;
2x² - 3x - 5 = 0
From the first equation the coefficient of x² is 1
Multiplying the coefficient of the x² with the constant -4, (1×-4) gives -4
The next thing that follows is having two factors of the -4 which when added gives -3 which is the coefficient of x
The two factors are -4 and +1
So the solution to it will be additive inverse of -4 and additive inverse of +1 which are +4 and-1
The second equation follows the same procedure but here we consider the coefficient of the x² because it's not 1 this time around
We multiply the coefficient by the constant which is 2 × -5 = -10
The two factors of -10 which when added gives the coefficient of x are -5 + 2 = -3
So the solution to the answer will now be additive inverse of -5 divided by the coefficient of x² which is 2 and the additive inverse of +2 divided by the coefficient of x² >> Ans, x = (+5)/2=2 1/2 or x = (-2)/2 = -1

Please note: This approach applies to the quadratic equation only factorization method can solve

By OJELABI OLUWASEYI( Bsc Industrial Chemistry)

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