Appropriate Question:
If one zero of the polynomial (a + 5)x² + 13x + 6a is reciprocal of the other, find the value of a.
Answer:
[tex]\boxed{\bf\:a = 1 \: } \\ [/tex]
Explanation:
Given that, one zero of the polynomial (a + 5)x² + 13x + 6a is the reciprocal of the other.
[tex]\sf \: Let\: assume \:that\: \alpha \: and \: \dfrac{1}{ \alpha } \: be \: zeroes \: of \: (a + 5) {x}^{2} + 13x + 6a \\ [/tex]
We know,
[tex]\boxed{{\sf Product\ of\ the\ zeroes=\dfrac{Constant}{coefficient\ of\ x^{2}}}} \\ [/tex]
So, using this result, we get
[tex]\sf \: \alpha \times \dfrac{1}{ \alpha } \: = \: \dfrac{6}{a + 5} \\ [/tex]
[tex]\sf \:1 \: = \: \dfrac{6}{a + 5} \\ [/tex]
[tex]\sf\:a + 5 = 6 \\ [/tex]
[tex]\implies\sf \: \boxed{\bf \: a = 1 \: } \\ [/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]
