Question

if one zero of a polynomial 3x2-8x+2k+1 is seven times the other then find the value of k.
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Answer 1

Answer:

value of k is 7...........

Answer 2

$\;\;\underline{\textbf{\textsf{ Given:-}}}$

• The quadratic polynomial is _____

3x² - 8x + 2k + 1

• One zero of the given polynomial is 7 times the other.

$\;\;\underline{\textbf{\textsf{ To Find :-}}}$

• The value of " k "

$\;\;\underline{\textbf{\textsf{ Solution :-}}}$

Let one zero of the polynomial be α.

Then the other zero will be 7α.

$\underline{\:\textsf{ We know that :}}$

• $\sf Sum \: of \: the \: zeroes = - \dfrac{coefficient \: of \: x}{coefficient \: of \: {x}^{2} }$

• $\sf Product \: of \: zeroes = \dfrac{constant \: term}{coefficient \: of \: {x}^{2} }$

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$\underline{\:\textsf{ Now, substitute the given values :}}$

$\sf Sum \: of \: the \: zeroes = - \dfrac{coefficient \: of \: x}{coefficient \: of \: {x}^{2} }$

$\sf \longrightarrow \alpha + 7\alpha = -\dfrac{-8}{3} \\\\ \sf \longrightarrow 8\alpha = \dfrac{8}{3} \\\\ \sf \longrightarrow \alpha = \dfrac{1}{3}$

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$\underline{\:\textsf{Again :}}$

$\sf Product \: of \: zeroes = \dfrac{constant \: term}{coefficient \: of \: {x}^{2} } \\ \\ \sf \alpha \times 7 \alpha = \dfrac{2k + 1}{3} \\ \\ \sf \longrightarrow 7 \alpha {}^{2} = \dfrac{2k + 1}{3} \\ \\ \sf \longrightarrow 21( \dfrac{1}{3} ) {}^{2} = 2k + 1 \\ \\ \sf \longrightarrow 2k + 1 = 21 \times \dfrac{1}{3 \times 3} \\ \\ \sf \longrightarrow 2k = \dfrac{7}{3} - 1 \\ \\ \sf \longrightarrow k = \frac{7 - 3}{3 \times 2} \\ \\ \sf\longrightarrow k = \dfrac{4}{6} \\ \\ \sf \longrightarrow k = \dfrac{2}{3}$

$\;\;\underline{\textbf{\textsf{ Hence-}}}$

$\underline{\textsf{ The value k is \textbf{2/3 }}}.$

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