Question

@ganeshie8

Answer 1

but than you check this quadratic wrote above the discriminant is less than zero so have not real roots - yes ?

Answer 2

Yes

Answer 3

That means your textbook must be lying to you

Answer 4

I would agree, doesn't seem to be a true statement.

Answer 5

I think I should provide the complete statement

Answer 6

"If x = a+bi is a root of the equation f(x)=0 where f(x) is a polynomial with real coefficients then x = a-bi is also a root. So multiplying both gives us
x²-2ax+a²+b²
Thus a real quadratic factor of f(x)
This mean that polynomial f(x), with real coefficients can be expressed as a product of real factors. None of which is of degree greater than 2. Whether we can find these factors is another story."
Is this true?

Answer 7

It seems correct to me though

Answer 8

Yes, that is a different statement. Complex roots always come in conjugate pairs, so you can always express a polyomial as a product of quadratic and linear factors.

Answer 9

I have two more questions. Can you help on them?

Answer 10

Sure, ask

Answer 11

This is regarding the proof

If a polynomial equation has a complex root, they always occur in pair.

Proof:-
If x=a+bi then f(x) =c+di where c and d are real
Even powers of bi are real and odd are imaginary

So if the sign of b is changed, c remains same while sign of d changes
If a+bi is a root then
f(a+bi)= 0
c+di=0
c=0, d=0
c-di is also 0
Thus c-di is also 0
f(a-bi) = c-di
a-bi is also a root

Answer 12

Is the last assumption, that if c-di is 0 then a-bi also becomes root, correct?

Answer 13

f(a-bi) = c-di

If above statement equals 0, then f(a-bi) = 0. Clearly a-bi is a zero / root.

Answer 14

Your proof looks nice !

Answer 15

sorry that i dont understand it clearly but than a-bi mean root of f(x) so than why f(a-bi) not eqaul zero and wrote above that f(a-bi)= c-di ?

Answer 16

Yeah okay, I realize it was a silly question.
Now the main challenging problem

If the roots of the equation
\[\large\rm x^4-px^3+qx^2-pqx+1=0\ \]
are a,b,c and d, show that
\[\large\rm (a+b+c)(a+b+d)(b+c+d)(a+c+d)=1\ \]

Answer 17

\(\color{#0cbb34}{\text{Originally Posted by}}\) @jhonyy9
sorry that i dont understand it clearly but than a-bi mean root of f(x) so than why f(a-bi) not eqaul zero and wrote above that f(a-bi)= c-di ?
\(\color{#0cbb34}{\text{End of Quote}}\)
Because you're taking the supposed hypothesis to be true. We have to prove that a-bi also becomes a root. Although it is obviously true, but we actually have to show why it is true.
f(a-bi) is obviously going to give c-di
Now the main thing was to show that c-di =0

Answer 18

ohhh - ok. ty

Answer 19

Looks tricky! Could you post it as a separate question so that others also get a chance to try ?

Answer 20

Okay sure