Hi,

I need help with this equation, i must solve it by factorising this quadratic equation.

27x^2 - 2x = 8

Help appreciated.

Tim

2. Hello, Tim!

Is there a typo?

Solve it by factoring: . $27x^2 - 2x \:= \:8$
The quadratic: . $27x^2 - 2x - 8 \:=\:0$ cannot be factored.

3. Hello Soroban,

Why ?

The discriminant is positive... Sure, the solution is quite ugly, but it remains factorisable.
Well this is certainly not the way to do it since it's asked to factorize first but it's false to state what you said...

4. Hello, Moo!

The discriminant is positive...
Sure, the solution is quite ugly, but it remains factorisable.
As soon as you mention "Discriminant", you are no longer factoring.
. . You are obviously invoking the Quadratic Formula.

But it's false to state what you said...
I don't agree . . .

According to my understanding of "factorable", $x^2-4x + 6$ doesn't factor.

Sure, we can use the Quadratic Formula and get: . $x \;=\;2 \pm i\sqrt{5}$

We can say that we have two factors: . $\left(x - [2 + i\sqrt{5}]\right)\,\left(x - [2-i\sqrt{5}]\right)$

. . but I would not claim that we factored it.

If we include irrational and complex expressions, everything factors.

. . $x^4 + 1 \;=\;\left(x-\sqrt{i}\right)\left(x + \sqrt{i}\right)\left(x - i\sqrt{i}\right)\left(x + i\sqrt{i}\right)$

. . $x + 4 \;=\;\left(\sqrt{x} - 2i\right)\left(\sqrt{x}+2i\right)$

5. Oh, I think we just don't agree with the term "factorable"...
There will always be a way to factor it

But I'm not sure to get the reason why it is not factorable in your terms...
Your example of x²-4x+6 shows that we factored it... since we got two factors... whatever the way we used is, isn't it ?

There is also the property :
if a trinomial is factorable, then, it can be written the following way : c(x-a)(x-b).

Huh ?

6. ok - back to the actual question - use your quadratic equation with a=27, b=-2, and c =-8