Polynomials Q (1 Viewer)

[Constip8edSkunk]

need help plz for just 1 more question in my last minute scramble for 4U assessment 2moro..

Prove that if the equation (x-b)(x-c)+(x-c)(x-a)+(x-a)(x-b)=0 has a double root, then a=b=c. (note that a, b, c are real numbers)

thanx

 

[spice girl]

Let P(x) = (x-b)(x-c)+(x-c)(x-a)+(x-a)(x-b)

then since P(a) = P(b) = P(c) = 0, then a, b, c are roots

if P(x) has a dbl root, and since it's a quadratic, then there is only one soln for which P(x) = 0

thus a=b=c

 

[Constip8edSkunk]

but P(a) P(b) and P(c) doesnt = 0

 

[spice girl]

heh...woops sorry

lemme do this again:

P(x) = (x-b)(x-c)+(x-c)(x-a)+(x-a)(x-b) = 3x^2 - 2(a+b+c)x + (ab + bc + ca)
P'(x) = 6x - 2(a + b + c)

If P has dbl roots, then there is one root (call it r) s.t. P(r) = P'(r) = 0

P'(r) = 0 implies r = (a+b+c)/3

then P(r) = 3{(a+b+c)/3}^2 - 2(a+b+c)*{(a+b+c)/3} + (ab + bc + ca)
= (-1/3)*(a+b+c)^2 + (ab + bc + ca)
= 0
thus (a+b+c)^2 = 3(ab+bc+ca)
(a^2 + b^2 + c^2 + 2(ab+bc+ca)) = 3(ab+bc+ca)
a^2 + b^2 + c^2 - (ab+bc+ca) = 0

(1/2)*{(a-b)^2 + (b-c)^2 + (c-a)^2} = 0
thus a=b=c

 

[ToRnaDo]

P(x) = (x-a)(x-b) + (x-c)(x-b) + (x-a)(x-c)
P'(x) = -a(x-b) - b(x-a) - c(x-b) - b(x-c) - a(x-c) - c(x-a)
and u solve P'(x)=0 and u will end up with the result
x= (ab+ac+bc)/(a+b+c)
then u sub x in the equation of P(x) to solve it =0

Well it will take a long time to solve it...I think u will end up with the result a=b=c.

Good luck for ur assesment tmr n for myself as well coz i have 2,3,4 Units Maths this week heheh

 

[Constip8edSkunk]

ok thanks for your helps... and i definitely need the luck lol, gluck 2 u 2