Question

Find zeros of following polynomial and verify it's relationship between zeros and coefficient.
$({7y}^{2}) - (\frac{11y}{3})- (\frac{2}{3}) = 0$
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Answer 1

Answer :-

$\to \sf \: Zeros \: are \: \: \frac{ - 1}{7} \: \: and \: \: \frac{2}{3}$

Explanation :-

We have

$\implies \sf{({7y}^{2}) - (\frac{11y}{3})- (\frac{2}{3}) = 0} \: \\ \\ \implies \sf{\frac{21 {y}^{2} - 11y - 2}{3} = 0} \\ \\ \implies \sf{21 {y}^{2} - 11y - 2 = 0} \: ....eq.(1) \\ \\ \bf{ \green{ split \: the} \red{ \: middle \: term \: }} \\ \\ \implies \sf{21 {y}^{2} - (14 - 3)y - 2 = 0} \\ \\ \implies \sf{21 {y}^{2} - 14y + 3y - 2 = 0} \\ \\ \implies \sf{ 7y(3y - 2) + 1(3y - 2) = 0}$

$\implies \sf{ (3y - 2)(7y + 1) = 0} \\ \\ \: (1) \bf \: if \: \\ \to \sf{ 3y - 2 = 0 }\\ \\ \to \sf{3y = 2} \\ \\ \to \boxed{\sf{y = \frac{2}{3} }} \\ \\ (2) \bf \: if \: \\ \to \sf{ 7y + 1 = 0} \\ \\ \to \boxed{ \sf{y = \frac{ - 1}{7} }}$

Verification :-

$(1) \sf{ \: if \: \: y = \frac{2}{3} } \: \: \: put \: in \: eq.(1) \\ \\ \to \: 21 { \bigg( \frac{2}{3} \bigg) }^{2} - 11 \times \frac{2}{3} - 2 = 0 \\ \\ \to \: \frac{84}{9} - \frac{22}{3} - 2 = 0 \\ \\ \to \: \frac{84 - 66 - 18}{9} = 0 \\ \\ \to \: \frac{0}{9} = 0 \\ \\ \to \: \boxed{0 = 0}$

$(2) \bf \: if \: \: y = \frac{ - 1}{7} \\ \\ \to \: 21 \times \frac{1}{49} - \frac{11 \times( - 1) }{7} - 2 = 0 \\ \\ \to \: \frac{21}{49} + \frac{11}{7} - 2 = 0 \\ \\ \to \: \frac{21 + 77 - 98}{49} = 0 \\ \\ \to \boxed{ 0 = 0}$

Hence verified.

$\star \sf{ \: \: Coefficient \: of \: {y}^{2} =7} \: \\ \\ \star \: \sf{ Coefficient \: of \: y = - \frac{11}{3} }$