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

Answer 1

If A and B are roots of 2x^2 - 7x + k, then on comparing with ax^2 + bx + c, a = 2, b = - 7, c = k

Sum of roots = A + B = - b/a = 7/2

Product of roots = AB = c/a = k/2

Given, A² + B² + AB = 67/4

=> A² + B² + 2AB - AB = 67/4

=> (A + B)² - AB = 67/4

=> (7/2)² - k/2 = 67/4

=> 49/4 - k/2 = 67/4

=> - 18/4 = k/2

=> - 9 = k

Answer 2

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

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