Question

What is the product of zeroes of Polynomial P(x) = x2 – 9x + 10​

Answer 1

$\textbf{Given:}$

$\textsf{Polynomial is}\;\mathsf{x^2-9x+10}$

$\textbf{To find:}$

$\textsf{Product of zeroes of the given polynomial}$

$\textbf{Solution:}$

$\textbf{Concept used:}$

$\boxed{\begin{minipage}{7cm} \mathsf{For\;the\;polynomial\;ax^2+bx+c\;}\\\\\mathsf{Sum\;of\;zeroes=\dfrac{-b}{a}}\\\\\mathsf{Product\;of\;zeroes=\dfrac{c}{a}} \end{minipage}}$

$\mathsf{Consider,\;x^2-9x+10}$

$\mathsf{Product\;of\;zeroes=\dfrac{c}{a}}$

$\mathsf{Product\;of\;zeroes=\dfrac{10}{1}}$

$\implies\boxed{\mathsf{Product\;of\;zeroes=10}}$

$\underline{\mathsf{Verification:}}$

$\mathsf{x^2-9x+10}$

$\mathsf{=(x^2-9x+\frac{81}{4})-\frac{81}{4}+10}$

$\mathsf{=(x-\frac{9}{2})^2-\frac{41}{4}}$

$\mathsf{=(x-\frac{9}{2})^2-(\frac{\sqrt{41}}{2})^2}$

$\mathsf{=(x-\frac{9}{2}-\frac{\sqrt{41}}{2})(x-\frac{9}{2}+\frac{\sqrt{41}}{2})}$

$\mathsf{=(x-(\frac{9}{2}+\frac{\sqrt{41}}{2}))(x-(\frac{9}{2}-\frac{\sqrt{41}}{2}))}$

$\implies\mathsf{Zeroes\;are\;\frac{9}{2}+\frac{\sqrt{41}}{2}\;\&\;\frac{9}{2}-\frac{\sqrt{41}}{2}}$

$\mathsf{Product\;of\;zeros=(\frac{9}{2}+\frac{\sqrt{41}}{2})(\frac{9}{2}-\frac{\sqrt{41}}{2})}$

$\mathsf{=\frac{81}{4}-\frac{41}{4}}$

$\mathsf{=\frac{81-41}{4}}$

$\mathsf{=\frac{40}{4}}$

$\mathsf{=10}$

$\textbf{Find more:}}$

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