How To Find The Factors Of THis Quadratic Equation?

[optics.tech]

Hi everyone,

Can someone please tell me how to find the factors of the below quadratic equation?

$$3x^2 - 9x + 4 = 0$$

I had already tried to find them by using the below method and wasn't able to continue further because of I do not understand.

Huygen

$$(3x^2 - 7x + 4) -2x = 0$$
$$(3x-4)(x-1)-2x=0$$

 

[DonAntonio]

Hi everyone,

Can someone please tell me how to find the factors of the below quadratic equation?

$$3x^2 - 9x + 4 = 0$$

I had already tried to find them by using the below method and wasn't able to continue further because of I do not understand.

Huygen

$$(3x^2 - 7x + 4) -2x = 0$$
$$(3x-4)(x-1)-2x=0$$

Since $\,\Delta:=b^2-4ac=9^2-4\cdot 3\cdot 4=33\,$ , this quadratic's roots are ugly:
$$x_{1,2}=\frac{-b\pm\sqrt\Delta}{2a}=\frac{9\pm\sqrt{33}}{6}$$ , so we can now factor
$$3x^2-9x+4=3\left[x-\left(\frac{9-\sqrt{33}}{6}\right)\right]\left[x-\left(\frac{9+\sqrt{33}}{6}\right)\right]$$
Ugly, indeed.

DonAntonio

 

[optics.tech]

Since $\,\Delta:=b^2-4ac=9^2-4\cdot 3\cdot 4=33\,$ , this quadratic's roots are ugly:
$$x_{1,2}=\frac{-b\pm\sqrt\Delta}{2a}=\frac{9\pm\sqrt{33}}{6}$$ , so we can now factor
$$3x^2-9x+4=3\left[x-\left(\frac{9-\sqrt{33}}{6}\right)\right]\left[x-\left(\frac{9+\sqrt{33}}{6}\right)\right]$$
Ugly, indeed.

DonAntonio

No, not this method.

There is available another method than this one.

 

[D H]

Completing the squares.

 

[eumyang]

Hi everyone,

Can someone please tell me how to find the factors of the below quadratic equation?

$$3x^2 - 9x + 4 = 0$$

Are you looking for the factors, or are you looking for the roots? If you are looking for factors, and they have to have integer coefficients, then you won't find any, as DonAntonio has shown. If you don't want to use the quadratic formula (again, as DonAntonio has shown), you'll need to complete the square, which D H suggested.

 

[Mentallic]

In comparison, if you use the quadratic formula on the separate quadratic you factorized:

$$(3x^2 - 7x + 4) -2x = 0$$
$$(3x-4)(x-1)-2x=0$$

So we're looking at $3x^2-7x+4$ only, then the discriminant

$$\Delta=b^2-4ac$$$$=(-7)^2-4(3)(4)$$$$=49-48=1$$

and thus since the square root of that is a rational number, then you can factorize the quadratic using only integer coefficients as you've shown.

 

[optics.tech]

Completing the squares.

Yes, you are correct.