if alpha and beta are zeroes of polynomial p(x)=2x2-7x+5. Find a polynomial whose zeroes are 2alpha+1 and 2beta+3
Answer :
Given : and are two zeros of the polynomial
Equation f ( x ) = 2 x2- 7 x + 5
So,
We know from relationship between zeros and coefficient ,
Sum of zeros = , So
+ = --- ( 1 )
And
Product of zeros = , So
= ------ ( 2 )
Taking Whole square of equation 1 , we get
( + )2 =
And
( - )2 + 4 = , Substitute value from equation 2 , we get
( - )2 + 4 ( ) =
( - )2 + 10 =
( - )2 = - 10
( - )2 =
( - )2 =
- = ---- ( 3 )
Add equation 1 and 3 , We get
2 = 5
= , Substitute that value in equation 3 we get
- = , So
= 1
So,
We get = and = 1
Then
2 + 1 = 2 ( ) + 1 =5 + 1 = 6
And
2 + 3 = 2 ( 1 ) + 3 = 2 + 3 = 5
Then
Sum of zeros of required polynomial = (2 + 1 )+ (2 + 3 ) = 6 + 5 = 11
And
Product of zeros of required polynomial = (2 + 1 )(2 + 3 )= ( 6 ) ( 5 ) = 30
And we know formula for polynomial when some of zeros and product of zeros we know :
Polynomial = k [ x2 - ( Sum of zeros ) x + ( Product of zeros ) ] , Here k is any non zero real number.
Substitute values , we get
Polynomial = k [ x2 - ( 11 ) x + ( 30 ) ]
Polynomial = k [ x2 - 11 x + 30]
Then
Quadratic polynomial = k [ x2 - 11 x + 30] = x2 - 11 x + 30 [on taking k = 1] ( Ans )
Given : and are two zeros of the polynomial
Equation f ( x ) = 2 x2- 7 x + 5
So,
We know from relationship between zeros and coefficient ,
Sum of zeros = , So
+ = --- ( 1 )
And
Product of zeros = , So
= ------ ( 2 )
Taking Whole square of equation 1 , we get
( + )2 =
And
( - )2 + 4 = , Substitute value from equation 2 , we get
( - )2 + 4 ( ) =
( - )2 + 10 =
( - )2 = - 10
( - )2 =
( - )2 =
- = ---- ( 3 )
Add equation 1 and 3 , We get
2 = 5
= , Substitute that value in equation 3 we get
- = , So
= 1
So,
We get = and = 1
Then
2 + 1 = 2 ( ) + 1 =5 + 1 = 6
And
2 + 3 = 2 ( 1 ) + 3 = 2 + 3 = 5
Then
Sum of zeros of required polynomial = (2 + 1 )+ (2 + 3 ) = 6 + 5 = 11
And
Product of zeros of required polynomial = (2 + 1 )(2 + 3 )= ( 6 ) ( 5 ) = 30
And we know formula for polynomial when some of zeros and product of zeros we know :
Polynomial = k [ x2 - ( Sum of zeros ) x + ( Product of zeros ) ] , Here k is any non zero real number.
Substitute values , we get
Polynomial = k [ x2 - ( 11 ) x + ( 30 ) ]
Polynomial = k [ x2 - 11 x + 30]
Then
Quadratic polynomial = k [ x2 - 11 x + 30] = x2 - 11 x + 30 [on taking k = 1] ( Ans )