Question

Find zeroes of the polynomial given and verify the relation.
X^2 -4root3x - 15

Answer 1

Answer:

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Step-by-step explanation:

$d = {b}^{2} - 4ac \\ d \: = {(4 \sqrt{3} )}^{2} - 4(1)( - 15) \\ d = 48 + 60 \\ d = 108 = 6 \sqrt{3} \\ \\ now \: apply \: quadratic \: formula \: and \: you \: will \: get \: zeroes \: as \\ x \: = 5 \sqrt{3} \: or \: x = - \sqrt{3} \\ \\ now \: we \: need \: to \: prove \: relation \\ sum \: of \: zeros \: = \frac{ - (coefficient \: of \: x)}{coefficient \: of \: {x}^{2} } \\ \\ now \: take \: lhs \: we \: get \\ sum \: of \: zeros \: = 5 \sqrt{3} + \sqrt{3} \\ sum \: of \: zeroes \: = = - 4 \sqrt{3} \\ \\ now \: take \: rhs \: \\ \frac{ - (cofficient \: of \: x)}{coefficient \: of \: {x}^{2} } = \frac{ - ( - 4 \sqrt{3)} }{1} = 4 \sqrt{3} \\ \\ \\ so \: we \: can \: say \: that \: this \: relation \: is \: true \: because \: lhs \: = rhs \\ \\ similary \: you \: can \: prove \: for \: product \: of \: zeros \: = \frac{constant \: term}{coefficient \: of \: {x}^{2} }$