rajat Rastogi

• Jul 1st, 2005

 

(x-b)*(x-c)=x2-11x+a-------(1) 
b+c=11 
bc=a 
(x-b)*(x-d)=x2-14x+2a.........(2) 
b+d=14 
bd=2a 
 
solving we get 
 
b=8 
c=3 
d=6 
which gives 
a=24 

Sana

• Aug 31st, 2007

 

Clearly the answer is 24.

If a=24;
In equation 1 : 24 = (-8) * (-3)  ;  -11 = (-8) + (-3)

2a+48;
In equation 2 : 48 = (-8) * (-6)   ;   -14 = (-8) + (-6)

Hence,a=24 with common factor (-8)

mohuas

 Profile Answers by mohuas 

• May 22nd, 2008

 

Let α is the common factor, β is a factor of 1st equation and γ is a factor of 2nd equation.
Then we can write that

α+β=11 -----(i)        αβ = a ----(ii)

α+γ=11 -----(i)        αγ = a ----(ii)

=> β = 3 and α = 8  => αβ = a = 24

prititripathi

 Profile Answers by prititripathi 

• Feb 20th, 2011

 

let 2 factor of eq. 1 a an b

so a+b=11 n ab=a

simmilarly for eq.2 ,2 factor let a n c

so a+c=14 n ac=2a

on solving we get 24

Tomer

• Sep 24th, 2011

 

x^2-11x+a=0 -eq1
x^2-14x+2a=0 - eq2 subtracting both eq
- + -
--------------------
0+3x-a=0 i.e 3x-a=0 this is x=a/3
now put the value of x in eq 1

(a/3)^2-11(a/3)+a=0
a=24

putting value of x in eq 2 we will again get a=24