Answer:
[tex]\boxed{\bf \:a = 3 \: } \\ [/tex]
Step-by-step explanation:
Given that, one zero of the polynomial (a²+ 9)x² + 13x + 6a is reciprocal of the other.
Let assume that
[tex]\sf \: \alpha \: and \: \dfrac{1}{ \alpha } \: be \: zeroes \: of \: f(x) = (a^2 + 9){x}^{2} + 13x + 6a \\ [/tex]
We know,
[tex]\boxed{{\sf Product\ of\ the\ zeroes=\dfrac{Constant}{coefficient\ of\ x^{2}}}} \\ [/tex]
So, on substituting the value, we get
[tex]\sf \: \alpha \times \dfrac{1}{ \alpha } \: = \: \dfrac{6a}{a^2 + 9} \\ [/tex]
[tex]\sf\:1 = \dfrac{6a}{ {a}^{2} + 9 } \\ [/tex]
[tex]\sf\:{6a} = { {a}^{2} + 9 } \\ [/tex]
[tex]\sf\: { {a}^{2} + 9 } - 6a = 0 \\ [/tex]
[tex]\sf\: {a}^{2} - 6a + 9 = 0 \\ [/tex]
[tex]\sf\: {a}^{2} - 3a - 3a + 9 = 0 \\ [/tex]
[tex]\sf\: a(a - 3) - 3(a - 3) = 0 \\ [/tex]
[tex]\sf\: (a - 3)(a - 3) = 0 \\ [/tex]
[tex]\sf\: (a - 3)^{2} = 0 \\ [/tex]
[tex]\sf\: a - 3 = 0 \\ [/tex]
[tex]\implies\sf\:a = 3 \\ [/tex]
Hence,
[tex]\implies\sf \: \boxed{\bf \: a = 3\: } \\ [/tex]
[tex]\rule{190pt}{2pt}[/tex]
Short Cut Trick:
If one zero of the polynomial ax² + bx + c is reciprocal of the other, then a = c
[tex]\rule{190pt}{2pt}[/tex]
Additional Information:
[tex]\sf \: If\: \alpha, \beta, \gamma \: are \: zeroes \: of \: {ax}^{3} + {bx}^{2} + cx + d, \: then \\ [/tex]
[tex]\qquad\sf \: \alpha + \beta + \gamma = - \dfrac{b}{a} \\ [/tex]
[tex]\qquad\sf \: \alpha \beta + \beta \gamma + \gamma \alpha = \dfrac{c}{a} \\ [/tex]
[tex]\qquad\sf \: \alpha \beta \gamma = - \dfrac{d}{a} [/tex]
