Question

if one of the zeros of the polynomial 3x^2-8x+2k+1 is seven times the other, then find the zeros and value of k

Answer 1

● THE GIVEN POLYNOMIAL IS

=> P(x) = 3x² - 8x + 2k + 1

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◢LET'S

$\alpha \: \: AND \: \: \beta \: \: ARE \: \: THE \: \: ZEROES \: \: \\ \ OF \: \: THE \: \: GIVEN \: \: EQUATION \:$
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◢NOW,

● ACCORDING TO THE QUESTION

$= > \alpha = 7 \beta \: \: \: .....Eq _{1}$
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● WE KNOW THAT

$= > SUM \: OF \: THE \: ZEROS \: =[ \frac{ - (COEFFICENT \: \: OF \: \: x)}{(COEFFICIENT \: \: OF \: \: {x}^{2}) }$]

◢THEN,

$= > \alpha + \beta \: = \frac{ - ( - 8)}{3} = \frac{8}{3} \\ \\ = > 7 \beta + \beta = \frac{8}{3} \: \: \: .....(By \: \: Eq _{1} )\\ \\ = > 8 \beta = \frac{8}{3} \\ \\ = > \beta = \frac{1}{3}$

◢AND,

$= > \alpha = 7 \beta \: \: \: \: \: ....(By \: \: Eq _{1}) \\ \\ = > \alpha = 7 \times \frac{1}{3} \\ \\ = > \alpha = \frac{7}{3}$
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◢NOW,

$= > PRODUCT \: \: OF \: \: ZEROS \: \: = \frac{CONSTANT \: \: TERM}{COEFFICIENT \: \: OF \: \: {x}^{2} } \\ \\ = > \alpha . \beta = \frac{2k + 1}{3}$

● PLUG THE VLAUES

$= > \frac{7}{3} . \frac{1}{3} = \frac{2k + 1}{3} \\ \\ = > \frac{7}{9} = \frac{2k + 1}{3 } \\ \\ = > 2k + 1 = \frac{7}{3} \\ \\ = > 2k = \frac{7}{3} - \: 1 \\ \\ = > 2k = \frac{7 - 3}{3} = \frac{4}{3} \\ \\ = > k = \frac{2}{3}$
__________________[◢ANSWER]

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● THUS THE VALUE OF "K" is

$= > k = \frac{2}{3}$

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☆$BE \: \: BRAINLY$☆