If the zeros of x² – kx + 6 are in the ratio 3:2 Find k.

let the zeroes of x² – kx + 6 be α and β

Given that ratio is 3:2 hence

⇒ 2α = 3β

Compare x² – kx + 6 with standard form of quadratic equation ax² + bx + c

a = 1, b = –k and c = 6

Product of roots

⇒ αβ = 6

Substituting value of α from (i)

⇒ 3β² = 6 × 2

⇒ β² = 4

⇒ β = ±2

Substituting β = 2 in equation (i)

⇒ α = 3

Substituting β = –2 in equation (i)

⇒ α = –3

Hence, we have two values of α and β, (α, β) = (3, 2) and (–2, –3)

Consider α = 3 and β = 2

Sum of roots of quadratic

⇒ α + β = –(–k)

⇒ 3 + 2 = k

⇒ k = 5

Consider α = –3 and β = –2

Sum of roots of quadratic

⇒ α + β = –(–k)

⇒ (–3) + (–2) = k

⇒ k = –5

Hence k = ±5