Question

15.  If one zero of the polynomial (a2 – 9) x2 + 13x + 6a is reciprocal of the other, find the value of ‘a’.

Answer 1

Answer:

The value of a is either 7.243 or -1.243.

Step-by-step explanation:

The given equation is

$p(x)=(a^2-9)x^2+13x+6a$

If a polynomial is defined as

$p(x)=ax^2+bx+c$

then the product of its roots is c/a.

It is given that one zero of the given polynomial is reciprocal of the other.

Let the zeroes of the given polynomials are k and 1/k.

$k\times \frac{1}{k}=\frac{6a}{a^2-9}$

$1=\frac{6a}{a^2-9}$

$a^2-9=6a$

$a^2-9-6a=0$

Using quadratic formula

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

$x=\frac{-(-6)\pm\sqrt{(-6)^2-4(1)(-9)}}{2(1)}$

$x=7.243,-1.243$

Therefore the value of a is either 7.243 or -1.243.

Answer 2

Answer:

the value of a is 3 and your question is wrong as it should be a2 +9 instead of a2 – 9 .

Step-by-step explanation:

let the zeroes be α and 1/α

α*1/α= c/a

1= 6a/a2 + 9

6a=a2 +9

a2 -6a + 9=0

a2 -3a-3a+9=0(by middle term splitting)

a(a-3) - 3(a-3)=0

(a-3)=0 and (a-3)=o

therefore a=3