Complex numbers (1 Viewer)

[yeyjasminetime]

Hi!

Any hints to factorise 4x^2 -4x + 5??

Thank you x

 

[obliviousninja]

Hi!

Any hints to factorise 4x^2 -4x + 5??

Thank you x

Quadratic forumula, and if this is 4u complex numbers question, i presume you are gonna get negative discriminant. using (a+ib)^2, find the square root.

 

[yeyjasminetime]

Okay so the answer I got is

The actual answer is (2x-1+2i)(2x-1-2i)

Why?

 

[QZP]

Tell me how you got that answer and I can tell you where you went wrong.

 

[Drongoski]

You can work backwards if you like, or engineer a difference of 2 squares..

 

[obliviousninja]

 

[MrBeefJerky]

Did you use the quadratic formula? How come u dont have -b in front of the discriminant all over 2a (a=4)

 

[jyu]

complete the square
4x^2 -4x + 5=4x^2 -4x + 1 + 4= (2x-1)^2 - (2i)^2 etc.

 

[QZP]

View attachment 28765

You cannot do that.

 

[yeyjasminetime]

Don't know how I got stuck on something so simple haha and I was stuck for SO LONG ):

So thanks!

 

[obliviousninja]

Did you use the quadratic formula? How come u dont have -b in front of the discriminant all over 2a (a=4)

4x^2 -4x + 5 negative-negative

You cannot do that.

Why not?

 

[yeyjasminetime]

Guys it does work

 

[QZP]

Why not?

There is no equation to solve for x hence it would not make sense to use the quadratic formula. Though this is just being pedantic.

 

[obliviousninja]

There is no equation to solve for x hence it would not make sense to use the quadratic formula. Though this is just being pedantic.

Ask the OP, probs supposed to be a '...=0' at the end. Given that its a factorization question, thats where I went with my working, hence used quadratic.

 

[QZP]

Ask the OP, probs supposed to be a '...=0' at the end. Given that its a factorization question, thats where I went with my working, hence used quadratic.

I don't think you understand where I'm coming from :S To factorise an expression is completely okay. However, to the factorise an expression by converting it into an equation to apply the quadratic formula only to remove the equality in the end is utterly wrong. You see, the quadratic formula is derived from ax^2 + bx + c = 0 - it SOLVES for x. In an expression, there is nothing to solve for. I believe you could lose marks in the HSC for this so watch out! It's an invalid method in terms of concrete maths.

 

[hit patel]

Even I didnt get you. ^ sorry ^

 

[Sy123]

 

[QZP]

I think the key bit here is the second line, it shows the equivalence of ax^2 + bc + c = (factored form featuring roots). It didn't seem like a logical step for me to find the roots a, b of 4x^2 -4x + 5 = 0 only to say hence (x-a)(x-b) = 4x^2 -4x +5. Appreciate your help. Also, did you mean "therefore if we find the roots of " ", we can factor the expression"?

Sorry for all that were mislead. :/

 

[Sy123]

I think the key bit here is the second line, it shows the equivalence of ax^2 + bc + c = (factored form featuring roots). It didn't seem like a logical step for me to find the roots a, b of 4x^2 -4x + 5 = 0 only to say hence (x-a)(x-b) = 4x^2 -4x +5. Appreciate your help. Also, did you mean "therefore if we find the roots of " ", we can factor the expression"?

Sorry for all that were mislead. :/

Ah yes expression not equation haha

 

[RealiseNothing]

I don't think you understand where I'm coming from :S To factorise an expression is completely okay. However, to the factorise an expression by converting it into an equation to apply the quadratic formula only to remove the equality in the end is utterly wrong. You see, the quadratic formula is derived from ax^2 + bx + c = 0 - it SOLVES for x. In an expression, there is nothing to solve for. I believe you could lose marks in the HSC for this so watch out! It's an invalid method in terms of concrete maths.

It's totally valid.