Question

if one zero of the polynomial p(x)=(a^2+9)x^2+45x+6a is receprocal of other find the value of a​

Answer 1

Step-by-step explanation:

Factoring  a^2-6a+9 

The first term is,  a2  its coefficient is  1 .

The middle term is,  -6a  its coefficient is  -6 .

The last term, "the constant", is  +9 

Step-1 : Multiply the coefficient of the first term by the constant   1 • 9 = 9 

Step-2 : Find two factors of  9  whose sum equals the coefficient of the middle term, which is   -6 .

     -9   +   -1   =   -10

     -3   +   -3   =   -6   That's it

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above,  -3  and  -3 

                     a2 - 3a - 3a - 9

Step-4 : Add up the first 2 terms, pulling out like factors :

                    a • (a-3)

              Add up the last 2 terms, pulling out common factors :

                    3 • (a-3)

Step-5 : Add up the four terms of step 4 :

                    (a-3)  •  (a-3)

             Which is the desired factorization

therefore a= 3.

Answer 2

Answer:

$\large{\underline{\boxed{\sf a = 3}}}$

Step-by-step explanation:

Given polynomial;

P(x) = (a² + 9)x² + 45x + 6a

On comparing the given equation with ax² + bx + c, we get -

• a = (a² + 9)
• b = 45
• c = 6a

Also, the one zero is reciprocal to other,

Thus, if one zero is α, then other is 1/α.

$\sf product \: of \: zeroes = \frac{c}{a} \\ \\ \implies \sf \alpha \times \frac{1}{ \alpha } = \frac{6a}{ {a}^{2} + 9 } \\ \\ \implies \sf 1 = \frac{6a}{ {a}^{2} + 9} \\ \\ \implies \sf {a}^{2} + 9 = 6a \\ \\ \implies \sf {a}^{2} - 6a + 9 = 0$

On splitting the middle term ;

$\implies \sf {a}^{2} - 3a - 3a + 9 = 0 \\ \\ \implies \sf a(a - 3) - 3(a - 3) = 0 \\ \\ \implies \sf (a - 3)(a - 3) = 0 \\ \\ \implies \sf a = 3 \: \: or \: \: a = 3$

Hence, the value of a is 3.