If (x-b) is the HCF of f(x) = x^2 + ax + 1 and g(x) = x^2 - 3x + 2, find the value of a and b.

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12 Jun 2014 Solve

If (x-b) is the HCF of f(x) = x^2 + ax + 1 and g(x) = x^2 - 3x + 2, find the value of a and b.

OPtion

1) a = -2, b = 1
2) a = 2, b = -2
3) a = 1/2, b = 2
4) a = 3, b = 1
5) a = 3, b = -5/2
6) a = 4, b = -2
7) a = 1, b = 1
8) a = 1, b = -1
9) a = 5/2, b = -1
10) None of these

Solution

As (x-b) is the HCF of f(x) and g(x), (x-b) must be a factor of f(x) and g(x)g
Using factor theorem
Solve
Option 1)

Correct Option: 1 Best Solution (18)

Saadana Saravanan   10 years AGO

The factors of g(x) = x^2 - 3x + 2 are
(x-2) and (x-1). this implies the value of b will be either 2 or 1. Therefore it is enough to check the options which has b=2 or b=1.
i.e., options - 1) 3) 4) 7)
Out of these options, we can easily find only option 1) a = -2 , b = 1 satisfies the equation with factors (x-1)& (x-1).
HCF is (x-1).
Therefore option 1)

Like? Yes (1) | No   

Vishnu Prasad V   10 years AGO

Only for a=-2 && b=1 the above eqn satisfies the condition

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IFFAT IRAM   10 years AGO

f(x) = x^2 + ax + 1
g(x) = x^2 - 3x + 2
f(b) = b^2 + ab + 1 =0
g(b) = b^2 - 3b + 2=0
a=-2, b=1

Like? Yes (1) | No   

mansi garg   10 years AGO

X^2-3X+2=(x-1)(x-2)
since (x-b)is hcf so b can be 1 or 2
by taking b=1 we can say that x=-2

Like? Yes (1) | No   

Sundar Kumar   10 years AGO

consider x=1, a=-2, b=1
f(1)=1+(-2*1)+1 = 0
g(1)=1-(3*1)+2 =0 so HCF of (f(1),g(1))= 0
HCF= (x-b) = (1-1) = 0
Similarly,
f(2)= 4+(-2*2)+1= 1
g(2)= 4-(3*2)+2= 0
HCF of (1,0) = 1
(x-b)= (2-1)=1
Answer option 1) a=-2, b=1

Like? Yes (7) | No (7)  

Kunal Arora   10 years AGO

As (x-b) is a factor.
f(b) and g(b) should be zero.
Putting in f(x) and g(x).
we have..
b(a+3)=1
only (1) satisfies

Like? Yes (3) | No (4)  

krishn kumar gupta   10 years AGO

g(x)=x^2-3x+2
=(x-1)(x-2)
now by option 1
a=-2, b=1
then
f(x)=x^2-2x+1
=(x-1)(x-1)
therefore
H.C.F.=(x-1)
b=1 which satisfy the option 1.

Like? Yes (2) | No (3)  

Pargat Pal Singh   10 years AGO

a = -2, b = 1
f(x) = x^2 + ax + 1= x^2 -2x + 1= (x-1)*(x-1)
g(x) = x^2 - 3x + 2= (x-1)*(x-2)
HCF is (x-1)

Like? Yes (3) | No (5)  

Varunraj Reddy   10 years AGO

If a=-2
f(x)=x^2-2x+1=(x-1)(x-1)
g(x)=x^2-3x+2=(x-1)(x-2)

Then HCF becomes (x-1)
(x-b)=(x-1)
i.e.,b=1

Then a=-2,b=1
Ans:Option 1)

Like? Yes (2) | No (4)  

Prakash kumar   10 years AGO

for g(x), x= 2 or 1
now hit n trial

if b=1
a= -2 putting b=2 in eqn f(x)= 0
we get satisfy only for option 1)

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Vaishali Gupta   10 years AGO

Factor of x^2-3x+2 is (x-1)(x-2)
So b can be 1 or 2. For b=2 equ x^2+ax+1 can not be factored for corresponding a value.for a=-2 & b=1the factor (x-1) become hcf

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REHAN AHMAD   10 years AGO

x^2-3x+2=(x-1)(x-2)
if x=-2 then
equation will be x^2-2x+1=(x-1)^2
so this is satisfying the condition
so x-1 is the hcf for the above equation..
answer is 1..a=-2 b=1

Like? Yes (1) | No (3)  

Saraswathy   10 years AGO

x=b

Since x-b is the HCF, x=b is completely divisible by b^2+ab+1 nd b^2-3b+2
From (2), b=1 or b=2

Wen b = 1, a = -2
Wen b = 2, a = -5/2

Ans : a=-2, b=1 (Option 1)

Like? Yes (2) | No (4)  

Girjesh kumar   10 years AGO

if (x-b) is the HCF of f(x) and g(x) then x = b will satisfy both equation .
so b^2 + ab + 1 =0 and b^2 - 3b +2 = 0 ,
then first option is satisfying above equation.

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paaredi muthyam reddy   10 years AGO

substitute x=b in both the equations then solve them
then we will get the values of b as 1 and 2.

if b=1 then a will become -2
if b=2 then a will become -5/2


from given options 1 is correct

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CHETNA TUTEJA   10 years AGO

f (x)=x^2+ax+1
g (x)=x^2-3x+2
When a=-2
Then f (x)=x^2-2x+1 and g (x)=x^2-3x+2
On solving f (x) and g (x), we get
x-1=0
since (x-b) is the HCF of f (x) & g (x),
So we get b=1
Hence ans. Is (1) a=-2, b=1

Like? Yes (2) | No (5)  

Ramesh Sriram   10 years AGO

(x-b) is a factor so b is a root for this equation

substitute b in both equations

solve equations
then
a=-2 and b=1

Like? Yes (1) | No (4)  

MRITUNJAY KUMAR   10 years AGO

As (x-b)is factor of f(x) and g(x),then both will be divisible by (x-b).
after division we get,
b^2-3b+2=0 & b^2+ab+1=0.after solving,we get b=1&b=2,take b=1& we get a=-2.here we have to b=1 only bcz (x-1) is the only factor which is common.

Like? Yes (2) | No (5)