Question

If α and β are zeroes of the polynomial p(x)=2x²+5x+k such that α²+β²+αβ=21/4
Find value of 'k'​

Answer 1

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✦ Required Answer:

♦️ GiveN:

• $\large{\rm{\alpha}}$ and $\large{\rm{\beta}}$ are the zeroes of polynomial $\large{\rm{P(x) = 2{x}^{2}+5x+k}}$

• $\large{ \rm{ { \alpha }^{2} + { \beta }^{2} + \alpha \beta = \frac{21}{4} }}$

♦️ To FinD:

• Find the value of k

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✦ How to solve?

Before solving this question, we need to know the relationship of zeroes and coefficients of the quadratic polynomial. The relations are:

$\large{ \boxed{ \rm{ \purple{Sum \: of \: zeroes = - \frac{coefficient \: of \: x}{coefficient \: of \: {x}^{2} } }}}}$

And,

$\large{ \boxed{ \rm{ \purple{Product \: of \: zeroes = \frac{constant \: term}{coefficient \: of \: {x}^{2} } }}}}$

I think, we should always remember this relation. I know that people often use, sum of zeroes = -b/a and product of zeroes = c/a when the quadratic polynomial ax^2+bx+c, but this might cause confusion when the coefficients are changed.

So, let's solve this question,

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✦ Solution:

We have, $\large{\rm{\alpha}}$ and $\large{\rm{\beta}}$ are the zeroes of polynomial P(x).

Then, according to relation,

$\large{ \rm{ \longrightarrow \: \alpha + \beta = \frac{ - 5}{2} }} \\ \\ \large{ \rm{ \longrightarrow \: \alpha \beta = \frac{k}{2} }}$

Also given,

$\large{ \rm{ \longrightarrow \: { \alpha }^{2} + { \beta }^{2} + \alpha \beta = \frac{21}{4} }}$

Using identity,

$\large{ \rm{ \ast{(a + b)}^{2} = {a}^{2} + 2ab + {b}^{2} }} \\ \\ \large{ \rm{ \ast{(a + b) {}^{2} - ab = {a}^{2} + ab + {b}^{2} }}}$

Arranging $\large{\rm{\alpha}}$ and $\large{\rm{\beta}}$ equation accordingly,

$\large{ \rm{ \longrightarrow \: {( \alpha + \beta ) }^{2} - \alpha \beta = \frac{21}{4} }} \\ \\ \large{ \rm{ \longrightarrow \: ( \frac{ - 5}{2} ) {}^{2} - \frac{k}{2} = \frac{21}{4} }} \\ \\ \large{ \rm{ \longrightarrow \: \frac{25}{4} - \frac{k}{2} = \frac{21}{4} }} \\ \\ \large{ \rm{ \longrightarrow \: \frac{k}{2} = \cancel{ \frac{4}{4} }}} \\ \\ \large{ \rm{ \longrightarrow \: \frac{k}{2} = 1}} \\ \\ \large{ \rm{ \longrightarrow \: \boxed{ \green{ \rm{k = 2}}}}} \\ \\ \large{ \therefore{ \underline{ \underline{ \rm{ \red{Hence, \: solved \: \dag}}}}}}$

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Answer 2

$\purple{\underline{\pink{\boxed{\boxed{\red{\star{\sf Question:}}}}}}}$

$\rm{If \ \alpha \ and \ \beta \ are \ zeroes \ of \ the \ polynomial }$

$\rm{p(x) \ = \ 2{x}^{2}+5x+k \ such \ that }$

$\rm{{ \alpha}^{2}+{ \beta}^{2}+ \alpha \beta \ = \ \frac{21}{4}. \ Find \ the \ value \ of \ k}$

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$\purple{\underline{\orange{\boxed{\boxed{\green{\star{\sf Answer:}}}}}}}$

$\sf\large\underline\blue{Hence, \ the \ value \ of \ k \ is \ 2}$

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$\sf\large\underline\pink{Given:}$

$\rm{Given, \ quadratic \ polynomial \ is :}$

$\rm{p(x) \ = \ 2{x}^{2}+5x+k}$

$\rm{Also,}$

$\rm{{\alpha}^{2}+{ \beta}^{2}+ \alpha \beta \ = \ \frac{21}{4}}$

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$\sf\large\underline\blue{To \ Find}$

$\rm{The \ value \ of \ k }$

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$\green{\underline{\red{\boxed{\boxed{\purple{\star{\sf Solution:}}}}}}}$

$\rm{we \ are \ given \ that, }$

$\rm{\alpha \ and \ \beta \ are \ the \ zeroes }$

$\rm{of \ given \ polynomial}$

$\rm{Now , \ using \ relation}$

$\rm{\implies{ \alpha + \beta = \frac{ - 5}{2}} }$

$\rm{\implies{ \alpha \beta = \frac{k}{2} }}$

$\rm{but \ we \ are \ given,}$

$\rm\implies{ { \alpha }^{2} + { \beta }^{2} + \alpha \beta = \frac{21}{4} }$

$\rm\implies{{( \alpha + \beta ) }^{2} - \alpha \beta = \frac{21}{4} }$

$\rm{using}$

$\rm{ {(a + b)}^{2} = {a}^{2} + 2ab + {b}^{2} }$

$\rm{ and}$

$\rm{ {(a + b)}^{2} - ab = {a}^{2} + ab + {b}^{2} }$

$\rm\implies{{ ( \frac{ - 5}{2} )} ^{2} - \frac{k}{2} = \frac{21}{4} }$

$\rm{\implies{ \frac{25}{4} - \frac{k}{2} = \frac{21}{4} }}$

$\rm{\implies{ \frac{k}{2} = \cancel{ \frac{4}{4} }}}$

$\rm{\implies{ \frac{k}{2} = 1}}$

$\rm{\therefore{\boxed{\purple{\bf{k = 2}}}}}$

$\sf\large\underline\blue{Hence, \ the \ value \ of \ k \ is \ 2}$

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$\bf\pink{Hope \ it \ helps \ :)) }$