Question

If α, β are the zeroes of the quadratic polynomial x² + kx = 12, such that α – β = 1, find the value of k.

Answer 1

The given quadratic polynomial is x² + kx = 12 and α – β = 1

If we rearrange the polynomial then we get

p(x) = x² + kx – 12

We have,

Now, if we recall the identities

(a + b)² = a² + b² + 2ab

Using the identity, we get

(α + β)² = α² + β² + 2 αβ

( – k)² = α² + β² + 2( – 12)

⇒ α² + β² = k² + 24 …(3)

Again, using the identity

(a – b)² = a² + b² – 2ab

Using the identity, we get (α – β)² = α² + β² – 2 αβ

(1)² = α² + β² – 2( – 12) {∵ (α – β) = 1}

⇒ α² + β² = 1 – 24

⇒ α² + β² = – 23 …(4)

From eqⁿ (3) and (4), we get

k² + 24 = – 23

⇒ k² = – 23 – 24

⇒ k² = – 47

Now the square can never be negative, so the value of k is imaginary.