Question

If one zero of the polynomial (a2 + 1)x2 – 29x + (3a + 1) is reciprocal of the other, then sum of all possible values of a is

Answer 1

Answer:

Correct option is A)

We know that equation ax2+bx+c=0

Then sum of roots =a−b and product of roots=ac

Let the other zero be α

Therefore, the other zero is α1

Now, α×α1=a2−96a

=>1=a2−96a

=>a2+9−6a=0

=>a2−6a+9=0

=>a2−3a−3a+9=0

=>a(a−3)−3(a−3)=0

=>(a−3)(a−3)=0

=>a=3 and a=3

Answer 2

Answer:

The answer is 3.

Step-by-step explanation:

The given polynomial is , ( a² + 1 )x² - 29x + ( 3a + 1 )

                                         = x² + ( -29 )/( a² + 1 )x + ( 3a + 1 )/( a² + 1 )

It is a 2nd degree quadratic polynomial.

Two zeroes of the given polynomial are reciprocal to each other.

Let , the zeroes be α and β .

then , β = 1/α  ⇒ αβ = 1

Then , the general form of 2nd degree quadratic polynomials is ,

[if α , β be two zeroes ]

x² + ( α + β )x + αβ

Hence , αβ = ( 3a + 1 )/( a² + 1 )

⇒ ( 3a + 1 )/( a² + 1 ) = 1

⇒ 3a + 1 = a² + 1

⇒ a² - 3a = 0

⇒ a( a - 3 ) = 0

⇒ a = 0 , 3

the sum of the values of a is 3.