• YOU can help the next generation of students in the community!
    Share your trial papers and notes on our Notes & Resources page

Proving-polynomials (1 Viewer)

shaon0

...
Joined
Mar 26, 2008
Messages
2,029
Location
Guess
Gender
Male
HSC
2009
Prove that the roots of the equation kx(x+4)=(x-1)(x+2) are always real.
I cannot prove this and need help. All help will be appreciated :)
I keep getting unreal values in this.
 

Iruka

Member
Joined
Jan 25, 2006
Messages
544
Gender
Undisclosed
HSC
N/A
Work out the discriminant, which is a quadratic in k. Then show that this quadratic is positive definite.
 

shaon0

...
Joined
Mar 26, 2008
Messages
2,029
Location
Guess
Gender
Male
HSC
2009
Iruka said:
Work out the discriminant, which is a quadratic in k. Then show that this quadratic is positive definite.
lol...i still keep getting unreal values for the discriminant.
 

Iruka

Member
Joined
Jan 25, 2006
Messages
544
Gender
Undisclosed
HSC
N/A
OK, so kx(x+4)=(x-1)(x+2)

Then (k-1)x^2 + (4k-1)x +2=0
so the discriminant is (4k-1)^2 - 8(k-1)= 16k^2 -16k +9.

If we complete the square, we get

16k^2 -16k +9 = (4k - 2)^2 +5 > 0 for all k.
 

tommykins

i am number -e^i*pi
Joined
Feb 18, 2007
Messages
5,730
Gender
Male
HSC
2008
回复: Proving-polynomials

shaon0 said:
Prove that the roots of the equation kx(x+4)=(x-1)(x+2) are always real.
I cannot prove this and need help. All help will be appreciated :)
I keep getting unreal values in this.
kx^2 + 4kx = x^2 + x - 2

(k-1)x^2 + (4k-1)x - 2 = 0

Discriminant

(4k-1)^2 - 8.(k-1)

= 16k^2 - 8k + 1 - 8k + 8
= 16k^2 - 16k + 9
= (4k-2)^2 + 5

(4k-2)^2 + 5 is always > 0 as (4k-2)^2 > 0 and 5 > 0

Thus polynomial always has real roots.
 

shaon0

...
Joined
Mar 26, 2008
Messages
2,029
Location
Guess
Gender
Male
HSC
2009
Re: 回复: Proving-polynomials

thanks guys for your help. really appreciate it.
Thanks :)
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top