please answer this question tennetiraj
Answers
Option C
Step-by-step explanation:
Solution :-
Given quardratic polynomial P(x) = x^2+bx+1
On Comparing this with the standard quadratic Polynomial ax^2+bx+c
We have
a = 1
b= b
c = 1
Given that
The nature of the zeroes = Both are real
We have
If the discriminant is greater than or equal to zero then the zeroes are real .
D = b^2-4ac ≥ 0
=> b^2-4(1)(1)≥0
=> b^2 -4 ≥0
=> b^2 ≥4
Let b^2 = 4
=b=±√4
=> b = ±2
if b^2 > 4
=> b>±√4
=> b >±2
=> -2<b<2
The possible values of b are equal to -2 and 2
Answer:-
b has at most two distinct values
option (c)
Check:-
Suppose b= 2 then the Polynomial = x^2+2x+1
=> x^2+x+x+1
=> x(x+1)+1(x+1)
=> (x+1)(x+1)
To get the zeores we write P(x)=0
=> (x+1)(x+1) = 0
=> x+1 = 0
=> x = -1
The zeores = -1 and -1 which are real
Both are real
If b = -2 then
x^2-2x+1
=> x^2-x-x+1
=>x(x-1)-1(x-1)
=>(x-1)(x-1)
To get the zeores we write P(x)=0
=> x-1=0
=> x = 1
Zeroes are 1 and 1 which are real
If b = 1 then
x^+x+1 = 0
The discriminant = 1^2-4(1)(1)
=> 1-4
=> -3 <0
The zeroes are not real.
Used formulae:-
- The standard quadratic Polynomial ax^2+bx+c
- The discriminant of ax^2+bx+c is D= b^2-4ac
- If D> 0 ,it has real and distinct roots
- If D< 0 it has no real roots
- If D= 0 it has equal and real roots