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

ANSWER.

• The value of k is -9.

SOLUTION.

Given equation –

→ 2x² - 7x + k = 0

Comparing with ax² + bx + c = 0, we get,

→ a = 2

→ b = -7

→ c = k

Let α and β be the zeroes. Therefore,

→ α + β = -b/a = -(-7)/2 = 7/2

→ αβ = c/a = k/2

Here, it's given that,

→ α² + β² + αβ = 67/4

Adding αβ to both sides, we get,

→ α² + β² + 2αβ = 67/4 + αβ

→ (α + β)² - αβ = 67/4

→ (7/2)² - k/2 = 67/4

→ (49 - 2k)/4 = 67/4

Multiplying both sides by 4, we get,

→ 49 - 2k = 67

→ 2k = 49 - 67

→ 2k = -18

→ k = -9

★ So, the value of k is -9.

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