Question

Find the zeroes of the polynomial √2x 2 + 7x + 5√2 and verify the relationship between its zeroes and coefficients.

Answer 1

$Given \: polynomial \: \sqrt{2}x^{2}+7x+5\sqrt{2}$

/* Splitting the middle term, we get */

$=\sqrt{2} x^{2} + 2x + 5x + 5\sqrt{2}$

$= \sqrt{2} x^{2} +\sqrt{2} \times \sqrt{2}x + 5x + 5\sqrt{2}$

$= \sqrt{2}x( x + \sqrt{2}) + 5(x + \sqrt{2})$

$= (x+\sqrt{2})(\sqrt{2}x + 5 )$

$So, the \: value \: of \::\sqrt{2}x^{2}+7x+5\sqrt{2}$

$is \: zero \: when \: x+\sqrt{2}= 0 \:Or \: \sqrt{2}x + 5 =0$

$i.e., when \: x = -\sqrt{2} \:Or \: x = \frac{-5}{\sqrt{2}}$

$\therefore The \: zeroes \:of \: \sqrt{2} x^{2} + 2x + 5x + 5\sqrt{2}\:are \: -\sqrt{2} \:and \: x = \frac{-5}{\sqrt{2}}$

$Now, \blue{i) the \: sum \:zeroes \:of \:\sqrt{2}x^{2}+7x+5\sqrt{2} }$

$= - \sqrt{2} + \Big( \frac{-5}{\sqrt{2}}\Big)$

$= - \sqrt{2} - \Big( \frac{5}{\sqrt{2}}\Big)$

$= \frac{-(\sqrt{2})^{2} - 5)}{\sqrt{2}}$

$= \frac{- 2 -5}{\sqrt{2}}$

$= \frac{- 7}{\sqrt{2}}$

$\green {= \frac{- Coefficient \:of \: x }{coefficient \:of \:x^{2}}}$

$\blue {ii) the \: Product \:zeroes \:of \:\sqrt{2}x^{2}+7x+5\sqrt{2}}$

$= (- \sqrt{2}) \times \Big( \frac{-5}{\sqrt{2}}\Big)$

$=5$

$= \frac{5}{1}$

$\green { = \frac{ Constant \: term }{Coefficient \: of \:x^{2}}}$

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