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

Given :-

2x² - 7x + k

α² + β² + αβ = 67/4

Solution :-

We know that

Sum of zeroes = -b/a

We have

b = -7

a = 2

Sum of zeroes = -(-7)/2

Sum of zeroes = 7/2

Product of zeroes = c/a

c = k

a = 2

Product of zeroes = k/2

α² + β² + αβ = 67/4

α² + β² + (2αβ - αβ) = 67/4

Taking 2 as common

(α + β)² - αβ = 67/4

Putting all values

(7/2)² - (k/2) = 67/4

(7²/2²) - k/2 = 67/4

49/4 - k/2 = 67/4

49 - 2k/4 = 67/4

49 - 2k = 67

-2k = 67 - 49

-2k = 18

k = 18/-2

k = -18/2

k = -9/1

k = -9

Answer 2

Given : $\alpha,~\beta$ are the zeroes of the polynomial $\sf 2x^2-7x+k$, such that $\sf\alpha^2+\beta^2+\alpha\beta=\dfrac{67}{4}.$

To find : The value of k.

$\qquad$_______________

On comparing $\sf2x^2-7x+k$ to $\sf ax^2+bx+c,$ we get the values as :

$\:$

$\qquad\dashrightarrow~\sf a=2$

$\qquad\dashrightarrow \: \sf b = - 7$

$\qquad \dashrightarrow \: \sf c = k$

$\:$

☀️ We know that :

$\:$

★ Product of zeroes $\alpha\beta$ = $\sf \dfrac{c}{a}$

$\qquad \dashrightarrow \: \alpha \beta = \sf \dfrac{k}{2}$

$\:$

★ Sum of zeroes $\alpha+\beta$ = $\sf\dfrac{-b}{a}$

$\sf \qquad \dashrightarrow \: \alpha + \beta = \dfrac{ - ( - 7)}{2}$

$\sf \qquad \dashrightarrow \: \alpha + \beta = \dfrac{7}{2}$

$\:$

☀️ According to the question :

$\:$

$\sf \qquad \dashrightarrow \: { \alpha }^{2} + { \beta }^{2} + \alpha \beta = \dfrac{67}{4}$

$\:$

☀️ Here, squaring on both sides :

$\:$

$\sf \qquad \dashrightarrow \: { \bigg( \alpha + \beta \bigg)}^{2} = { \bigg \lgroup \dfrac{7}{2} \bigg \rgroup}^{2}$

$\sf \qquad \dashrightarrow \: { \alpha }^{2} + { \beta }^{2} + 2 \alpha \beta = \bigg \lgroup \dfrac{49}{4} \bigg \rgroup$

$\pmb{ \sf \qquad \dashrightarrow \: { \alpha }^{2} + { \beta }^{2} + \alpha \beta + \alpha \beta = \bigg \lgroup \dfrac{49}{4} \bigg \rgroup}$

$\:$

☀️ Substituting the values :

$\:$

$\sf \qquad \dashrightarrow \: \bigg \lgroup \dfrac{67}{4} \bigg \rgroup + \bigg \lgroup \dfrac{k}{2} \bigg \rgroup = \bigg \lgroup \dfrac{49}{4} \bigg \rgroup$

$\sf \qquad \dashrightarrow \: \bigg \lgroup \dfrac{k}{2} \bigg \rgroup = \dfrac{49 - 67}{4}$

$\sf \qquad \dashrightarrow \: \dfrac{k}{2} = \bigg \lgroup \cancel\dfrac{ - 18}{4} \: \bigg \rgroup$

$\sf \qquad \dashrightarrow \: \dfrac{k}{2} = \bigg \lgroup\dfrac{ - 9}{2} \bigg \rgroup$

$\pmb{ \sf \qquad \dashrightarrow \: k = - 9}$

$\:$

★ Therefore, the value of k is -9.