Quadratic polynomial problem

[rongmon]

P (x) is a quadratic polynomial such that P (1) = - P (2). If one root of the equation is –1, the other root is

(A) -4/5;

(B) 8/5;

(C) 4/5;

(D) -8/5.



Thanks.

[Dreyfus]

The other root lies b/w 1 nd 2 as polynomial changes sign from 1 to 2......so 8/5

[Homer Simpson]

1(B)

Vaibhav, i didnt quite understand your explanation. could you please elaborate? i used the sum and product of roots of quadratic equation to get 8/5.

[Dreyfus]

Tsuki...draw a graph of polynomial nd at x=1 if p(x) =t den at x=2 p(x)=-t now this means that b/w these two numbers p(x) changes its sign nd since px is polynomial hence continuous which implies that px when changed its sign from 1 to 2 den it must have become 0 somewhere b/w 1 nd 2

[Homer Simpson]

yeah, got it now! thank you :)

[ashutosh]

@vaibhav
Trick was good :D.  

@tsuki
Exact value can be found out by first taking general polynomial - x2+bx+c  

Put f(1) = -2f(2)
we get 3b+2c-5=0

and given alpha = -1
use alpha + beta = -b
  alpha. beta = c


You will get 2 equations solve for beta .. = 8/5

[rongmon]

I attempted solving using the general form as well.

Can you share the workings of your solution - i could not follow.

Thanks.

[Homer Simpson]

Rongmon, this is the working -

General form : ax^2 + bx + c

=> f(1) = -f(2)
=> a + b + c = -(4a + 2b + c)
=> 5a + 3b + 2c = 0
=> 5 + 3(b/a) + 2(c/a) = 0 -(i)

let roots be m,n and m = -1

Using sum and product of roots of quadratic equation -

     m+n = (-b/a) and mn = (c/a)
=>  n -1 = b/a    and  -n = c/a - substitute these in (i) and solve for n. you will get 8/5

[rongmon]

Thanks Tsuki.

[jack]

@ tsuki
why did you do m+n=(-b/a) instead of (b/a)?
please help

[Homer Simpson]

Jack, check this out-

http://www.mathwarehouse.com/quadratic/roots/formula-sum-product-of-roots.php