Question

if one zero of the polynomial p(x)=(a^2+9)x^2+13x+6a is the reciprocal of the other,find the value of a

Answer 1

Given

Polynomial p( x ) = ( a² + 9 )x² + 13x + 6a

One of the zero is reciprocal of the other

So, let the zeroes of the polynomial be t, 1 / t

Comparing the given polynomial with Ax² + Bx + C we get,

• A = a² + 9

• B = 13

• C = 6a

Product of zeroes = C / A

⇒ t × ( 1 / t ) = 6a / ( a² + 9 )

⇒ 1 = 6a / ( a² + 9 )

⇒ a² + 9 = 6a

⇒ a² - 6a + 9 = 0

Splitting the middle term

⇒ a² - 3a - 3a + 9 = 0

⇒ a ( a - 3 ) - 3 ( a - 3 ) = 0

⇒ ( a - 3 ) ( a - 3 ) = 0

⇒ a - 3 = 0 or a - 3 = 0

⇒ a = 3 or a = 3

Therefore the value of a is 3

Answer 2

Answer:

$\textsf{Let the Zeroes of Polynomial be x and \sf\frac{1}{x} }.$

$\sf p(x) =(a^2 + 9)x^2 + 13x + 6a \\\\ \sf\qquad we \:have\:: \\\\ \sf A=(a^2 + 9),\quad B =13,\quad C = 6a$

$\rule{170}{1}$

$\underline{\bf{\dag}\:\textsf{Product of Zeroes :}}$

$\dashrightarrow\tt\:\: Product_{zeroes}=\dfrac{Constant \:Term}{Coefficient \: of\: x^2}\\\\\\\dashrightarrow\tt\:\:x \times \dfrac{1}{x} = \dfrac{C}{A}\\\\\\\dashrightarrow\tt\:\:1 = \dfrac{6a}{(a^2 + 9)}\\\\\\\dashrightarrow\tt\:\:a^2 + 9 = 6a\\\\\\\dashrightarrow\tt\:\:a^2 + 9 - 6a = 0\\\\\\\dashrightarrow\tt\:\:(a)^2 + (3)^2 - (2 \times a \times 3) = 0\\\\\\\dashrightarrow\tt\:\:(a - 3)^2 = 0\\\\\\\dashrightarrow\tt\:\:(a - 3) = 0\\\\\\\dashrightarrow\:\:\underline{\boxed{\tt a = 3}}$

⠀

$\therefore\:\underline{\textsf{Therefore, value of a is equal to \textbf{3}}.}$