Question

Find the value of k, if the zeroes
of polynomial p(x) = (k-1) x² + 4x + (2k - 3)
are
reciprocals of each others

(a) -2

(b) 4 (c) 2 (d) -4​

Answer 1

Answer:

$\boxed{\sf \: (c) \: \: \: \: 2 \: }$

Step-by-step explanation:

Given that, the zeroes of polynomial p(x) = (k-1) x² + 4x + (2k - 3) are reciprocals of each other.

Let assume that

$\sf \: \alpha \: and \: \dfrac{1}{ \alpha } \: be \: the \: zeroes \: of \: p(x) = (k - 1) {x}^{2} + 4x + (2k - 3)$

We know,

$\boxed{{\sf Product\ of\ the\ zeroes=\dfrac{Constant}{coefficient\ of\ x^{2}}}}$

So, using this result, we get

$\sf \: \alpha \times \dfrac{1}{ \alpha } = \dfrac{2k - 3}{k - 1}$

$\sf \: 1 = \dfrac{2k - 3}{k - 1}$

$\sf \: 2k - 3 = k - 1$

$\sf \: 2k - k = 3 - 1$

$\implies\sf \: \sf \: k = 2$

$\rule{190pt}{2pt}$

Additional Information

$\begin{gathered} \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{{More \: identities}}}} \\ \\ \bigstar \: \bf{ {(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2} }\:\\ \\ \bigstar \: \bf{ {(x - y)}^{2} = {x}^{2} - 2xy + {y}^{2} }\:\\ \\ \bigstar \: \bf{ {x}^{2} - {y}^{2} = (x + y)(x - y)}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{2} - {(x - y)}^{2} = 4xy}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{2} + {(x - y)}^{2} = 2( {x}^{2} + {y}^{2})}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y)}\:\\ \\ \bigstar \: \bf{ {(x - y)}^{3} = {x}^{3} - {y}^{3} - 3xy(x - y) }\:\\ \\ \bigstar \: \bf{ {x}^{3} + {y}^{3} = (x + y)( {x}^{2} - xy + {y}^{2} )}\: \end{array} }}\end{gathered}\end{gathered}\end{gathered}$