Question

Can you use square roots to solve all quadratic equations?

Answer 1

If the question is about using the square root directly against the equation, the answer is definitely NO.

However, with certain transformation of a given equation into a different but equivalent form it is possible. Here is the idea.

Assume, for example, the same equation as analyzed in the previous answer:
`x^2+x=63`

If we could transform it to something like `y^2=b` then the square root of both sides would deliver a solution.
So, let's transform our equation to this form.
Expression `x^2+x` is not a square of anything, but `x^2+x+1/4` is a square of `x+1/2` because
`(x+1/2)^2=x^2+2*x*1/2+1/4=x^2+x+1/4`

Therefore, it is reasonable to transform the original equation into
`(x+1/2)^2-1/4=63` or
`(x+1/2)^2=253/4`
From the last equation, which is absolutely equivalent to the original one, using the operation of the square root, we derive two linear equations:
`x+1/2=sqrt(253)/2` and `x+1/2=-sqrt(253)/2`

So, two solutions are:
`x=(-1+sqrt(253))/2` and `x=(-1-sqrt(253))/2`

The above method is pretty universal and handy if you don't remember a formula for solutions of a quadratic equation. Let me illustrate this with another example.
`-3x^2+2x+8=0`

Step 1. Divide everything by `-3` to have `x^2` with a multiplier `1`:
`x^2-2/3x-8/3=0`

Step 2. Since a coefficient at `x` is `-2/3`, use `(x-1/3)^2` in a transformed equation:
`(x-1/3)^2-1/9-8/3=0` or
`(x-1/3)^2=25/9`

Step 3. Use square root:
`x-1/3=5/3` and `x-1/3=-5/3`

Step 4. Solutions:
`x=6/3=2` and `x=-4/3`