Question

Can every quadratic equation be solved by factorization or by splitting the middle term? When does one use quadratic formula?

Answer 1

No. Every quadratic equation has two solutions and can be factorized, but as level of difficulty rises, splitting may not be easy and one may tend to use quadratic formula.

Explanation:

According to fundamental theorem of algebra, a polynomial of degree `n` has `n` solutions. Hence a quadratic equation will always have two solutions. Factorization is one of the ways to solve such an equation. General process of factorization is as follows.

To factorize a quadratic polynomial of general form `ax^2+bx+c`, one should divide middle term `bx` in two parts, whose sum is `b` and product is `axxc`.

Assume factors of `ax^2+bx+c` to be `(lx+m)(px+q)` i.e.

`ax^2+bx+c=lpx^2+(lq+mp)x+mq`

So one splits `lpxxmq` into `lq` and `mp` and splitting of middle term to help factorization is key to the process.

However, splitting of numbers this way may not be easy. For example though

`3x^2-(root(3)6+root(3)36)x+2=(root(3)9x-root(3)2)(root(3)3x-root(3)4)`,

it may not be easy far every one to factorize `3x^2-(root(3)6+root(3)36)x+2`.

Generally, it is easier only when roots of the quadratic equation are rational numbers. This happens when the `a`, `b` and `c` are rational and discriminant `b^2-4ac` too is complete square of a rational number.

When we switch to irrational numbers or complex numbers, complexities grow and as according to fundamental theorem of algebra a quadratic equation has two roots, one can use quadratic formula. For example, using quadratic formula the factors of `ax^2+bx+c` will be given by

`(ax+(-b+sqrtb^2-4ac)/2)(x+(-b-sqrt(b^2-4ac))/(2a))`

Thus `ax^2+bx+c` can be factorized, the only difference is that one may not use splitting middle term, when complexities grow.

Hence, every quadratic equation has two solutions and can be factorized, but as level of difficulty rises, splitting may not be easy and one may tend to use quadratic formula.