Question

Please help me to find the answers...........If one of the polynomial (a×a+9)x×x+13x+6a is reciprocal of the other , find the value of 'a'​

Answer 1

Question

If one of the zeros of the polynomial (a² +9)x² + 13x + 6a is reciprocal of the other, find the value of 'a'.

Solution

Given polynomial is (a² + 9)x² + 13x + 6a.

Here, a = (a² + 9), b = 13 and c = 6a

We have to find the value of a.

Let us assume that the one zero of the polynomial be x.

Other zero is reciprocal of the first one i.e. of the x. So, other zero of the polynomial is 1/x.

We know that,

Product of zeros = c/a

Assumed zeros are x and 1/x and their product is 1.

→ (1/x) × (x) = 6a/(a² + 9)

→ 1 = 6a/(a² + 9)

Cross-multiply them

→ a² + 9 = 6a

→ a² + 9 - 6a = 0

→ a² - 6a + 9 = 0

Now, solve it by splitting the middle term

→ a² - 3a - 3a + 9 = 0

→ a(a - 3) -3(a - 3) = 0

→ (a - 3)(a - 3) = 0

On comparing we get,

→ a = 3

Answer 2

Given :-

• One zero of Polynomial (a² + 9)x² + 13x + 6a is reciprocal of the other ...

To Find :-

• find the value of 'a'. ?

Concept Used :-

The sum of the roots of the Equation ax² + bx + c = 0 , is given by = (-b/a)

and ,

→ Product of roots of the Equation is given by = c/a.

______________________

Solution :-

Comparing The Given Polynomial with ax² + bx + c = 0, we get,

→ a = (a² + 9)

→ b = 13

→ c = 6a

Now, Lets Assume That, One Root of The Equation is ɑ .

So, Other Root will be its Reciprocal = (1/ɑ).

So,

→ Product of Roots = (c/a) .

Putting Values :-

→ ɑ * (1/ɑ) = 6a/(a² + 9)

→ 1 * (a² + 9) = 6a

→ a² + 9 = 6a

→ a² - 6a + 9 = 0

Splitting The Middle Term Now,

→ a² - 3a - 3a + 9 = 0

→ a(a - 3) - 3(a - 3) = 0

→ (a - 3)(a - 3) = 0

→ (a - 3)² = 0

Putting Equal to Zero we get,

→ a = 3. (Ans).

Hence, The value of a will be 3.