Question

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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

Answer:

k = -9

Explanation:

Given :

α, β are the zeroes of the polynomial  2x² - 7x + k satisfying the condition

α² + β² + αβ = 67/4

To find :

the value of k

Solution :

For the given polynomial 2x² - 7x + k,

• x² coefficient = 2

• x coefficient = -7

• constant term = k

From the relation between zeroes and coefficients of a quadratic polynomial :

>> Sum of zeroes = -(x coefficient)/x² coefficient

>> Product of zeroes = constant term/x² coefficient

So,

α + β = -(-7)/2 = 7/2

αβ = k/2

We know,

(x + y)² = x²+ y² + 2xy

Similarly,

(α + β)² = α² + β² + 2αβ

(7/2)² = α² + β² + αβ + αβ

49/4 = (α² + β² + αβ) + k/2

49/4 = 67/4 + k/2  [ ∵ α² + β² + αβ = 67/4 ]

k/2 = 49/4 - 67/4

k/2 = (49 - 67)/4

k/2 = -18/4

k/2 = -9/2

k = -9

Therefore, the value of k is -9