Question

find the zero of the polynomial and verify the relationship between the zeros and the coefficient 4u²+8u​

Answer 1

Step-by-step explanation:

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Answer 2

$\bf{\underline{\underline{Given\: :}}}$

$\mathsf{\bigstar\: Polynomial = 4u^{2} + 8u}$

$\bf{\underline{\underline{To\: find\: :}}}$

$\mathsf{\bigstar\: Zeroes\:of\:the\:polynomial}$

$\bf{\underline{\underline{Solution\: :}}}$

$\mathsf{For\:zeroes\:of\:polynomial}$

$\mathtt{\implies\: 4u^{2} + 8u = 0}$

$\mathtt{\implies\: 4u(u + 2) = 0}$

$\mathtt{\implies\: 4u = 0 \: ; \: u + 2 = 0}$

$\mathtt{\implies\: u = \frac{0}{4} \: ; \: u = -2}$

$\mathtt{\implies\: u = 0 \: \: or \: \: -2}$

$\bf{\underline{\underline{Verification \: :}}}$

$\mathsf{\red{\diamond\: \: Sum\:of\:zeroes = \frac{-(x \:\: coeffiecient)}{x^{2}\: \:coefficient} }}$

$\mathtt{\rightarrow\: 0 + (-2) = \frac{-8}{4}}$

$\mathtt{\rightarrow\: -2 = -2}$

$\mathsf{\green{\diamond\: \: Product\:of\:zeroes = \frac{constant\:\:term}{x^{2}\:\: coefficient} }}$

∵ There is no constant term

∴ Constant term = 0

$\mathtt{\rightarrow\: 0 × (-2) = \frac{0}{4}}$

$\mathtt{\rightarrow\: 0 = 0}$

$\mathsf{In\:both\:the\:cases\:\: LHS = RHS}$

∴ Hence Proved