437 views

0 votes

f(x)=ax^{2}+bx+c

f(5)=25a+5b+c

f(2)=4a+2b+c

25a+5b+c = -3(4a+2b+c)

25a+5b+c = -12a -6b -3c

37a+11b+4c=0

f(5)=-3f(2)

3 is a root, hence

f(3)=0,

9a+3b+c=0

37a+11b+4c=0 -------(i)

9a+3b+c=0 ----------(ii)

On solving (i) and (ii) we get, b = a, c = -12a -----(iii)

Substituting (iii) in f(x) = 0

ax2 + ax – 12a = 0 => a(x2 + x – 12) = 0

Given that a ≠ 0 =>(x-3)(x+4) = 0 => x = 3 (or) -4

x=3 root is already given hence the other root is -4

but With the given two conditions (i) and (ii), it is not possible to find the value of a+b+c