Question

If α,β are the zeroes of the polynomial
2x2-7x +k satisfying the condition
α2+β2+αβ= 67/4, then find the value of k
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Answer 1

$\huge\mathcal{\fcolorbox{cyan}{black}{\pink{ANSWER}}}$

If α,β are the zeroes of the polynomial

$2 {x}^{2} - 7x + k$

$\therefore \: \alpha + \beta = - \frac{ - 7}{2} = \frac{7}{2} \\ \alpha \beta = k/2\:\:----(1)\\ { \alpha }^{2} + { \beta }^{2} = {( \alpha + \beta )}^{2} - 2 \alpha \beta \\ \: \: \: \: \: = {( \frac{7}{2} )}^{2} - 2k \\ = \frac{49}{4} - 2k/2 \\ = \frac{49 - 4k}{4} \: \: - - - - (2)$

Given,

${ \alpha }^{2} + { \beta }^{2} + \alpha \beta = \frac{67}{4} \\ \implies \: \frac{49 - 4k}{4} + k/2 = \frac{67}{4}\:\:\:\:\:\:[from\:eqn.\:\:(1)\:,\:(2)] \\ \implies \: \frac{49 - 4k + 2k}{4} = \frac{67}{4} \\ \implies \: 49 - 2k = 67 \\ \implies \: 2k = 67 + 49 = 116 \\ \implies \: k = \frac{116}{2} = 58 \\ \\ \therefore \: k = 58$

Answer 2

Given:-

$⇒ {2x}^{2} - 7x + k$

$⇒ {ɑ }^{2} + { β}^{2} + ɑβ = \frac{67}{4}$

Solution:-

$\text{we Know that}$

$⇒\text{sum of zeroes} = - \frac{ β}{ɑ }$

$\textbf{we have}$

$⇒β = - 7$

$⇒ɑ = 2$

$⇒\text{sum of zeroes} = - \frac{ - 7}{2}$

$⇒\text{sum of zeroes} = \frac{7}{2}$

$⇒\text{Product of zeroes} = \frac{c}{a}$

$⇒c = k$

$⇒a = 2$

$⇒\text{Product of zeroes} = \frac{k}{2}$

$⇒ {ɑ }^{2} + {β}^{2} + ɑβ = \frac{67}{4}$

$⇒ {ɑ }^{2} + { β}^{2} + (2ɑ β - ɑ β) = \frac{67}{4}$

$\text{Taking 2 as common}$

$⇒(ɑ + β {)}^{2} - ɑ β = \frac{67}{4}$

$\textbf{Putting all values}$

$⇒ { \left( \frac{7}{2} \right) \]}^{2} - ( \frac{k}{2} ) = \frac{67}{4}$

$⇒ \left( { \frac{7}{2} }^{2} \right) \] - \frac{k}{2} = \frac{67}{4}$

$⇒ \frac{49}{4} - \frac{k}{2} = \frac{67}{4}$

$⇒ 49 - 2k/4 = \frac{67}{4}$

$⇒49 - 2k = 67$

$⇒ - 2k = 67 - 49$

$⇒ - 2k = 18$

$⇒k = \frac{18}{ - 2}$

$⇒k = \frac{ - 18}{2}$

$⇒k = \frac{ - 9}{1}$

$⇒k = - 9$

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