Question

If one zero of the polynomial 3x2 – 8x - (2k +1) is seven times the other, find the zeroes of the polynomial and the value of k.

Answer 1

GIVEN :–

• One zero of the polynomial 3x² - 8x - (2k +1) = 0 is seven times the other.

TO FIND :–

• Value of "k" = ?

SOLUTION :–

• If one root of polynomial is 'α' then other is '7α'.

• We know that –

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$\bf \implies Sum \: \: of \: \: roots = - \dfrac{coefficient \: \: of \: \: x}{coefficient \: \: of \: \: {x}^{2}}$

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$\bf \implies \alpha + 7 \alpha = - \dfrac{( - 8)}{3}$

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$\bf \implies 8\alpha = \dfrac{8}{3}$

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$\bf \implies \large { \boxed{ \bf \alpha = \dfrac{1}{3}}}$

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• And –

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$\bf \implies Product \: \: of \: \: roots = \dfrac{constant\:\: term}{coefficient \: \: of \: \: {x}^{2}}$

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$\bf \implies ( \alpha )(7 \alpha )= \dfrac{ - (2k + 1)}{3}$

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$\bf \implies 7 {\alpha}^{2} = \dfrac{ - (2k + 1)}{3}$

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$\bf \implies 7 { \left( \dfrac{1}{3} \right)}^{2} = \dfrac{ - (2k + 1)}{3}$

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$\bf \implies 7 { \left( \dfrac{1}{3} \right)} = - (2k + 1)$

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$\bf \implies \dfrac{7}{3} = - (2k + 1)$

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$\bf \implies2k + 1 = - \dfrac{7}{3}$

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$\bf \implies2k = - 1- \dfrac{7}{3}$

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$\bf \implies2k = - \dfrac{10}{3}$

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$\bf \implies \large{ \boxed{ \bf k = - \dfrac{5}{3}}}$

Answer 2

Answer:

k = -5/3

Step-by-step explanation:

If one zero of the polynomial 3x² - 8x - (2k +1) is seven times the other.

Assume that the one zero is x. So, the other zero is 7x.

Sum of zeros = -b/a

In the given polynomial; a is 3, b is -8 and c is -(2k +1).

→ x + 7x = -(-8)/3

→ 8x = 8/3

→ x = 1/3

Product of zeros = c/a

→ x(7x) = -(2k+1)/3

→ 7x² = -(2k + 1)/3

→ 7(1/9) = -(2k + 1)/3

→ 7/3 = -2k - 1

→ 7 = -6k - 3

→ 10 = - 6k

→ k = -10/6

→ k = -5/3

Hence, the value of k is -5/3.