Question

Let p(x) = 2x⁴ – 3x3 – 5x² + 9x – 3 and two of its zeroes are √3 and −√3. Find the other two zeroes.

Answer 1

We have,

P(x) = 2x⁴ – 3x³ – 5x² + 9x – 3. 

Also,

Given that,

√3 and −√3 are zeros of given polynomial P(x).

Let the other two zeros are α and β.

∴ Sum of roots = α + β + √3 − √3

= \(\frac{-b}{a}\)

= \(\frac{-(-3)}{2}\) 

= \(\frac{3}{2}\).

⇒ α + β = \(\frac{3}{2}\).… (1)

And product of roots = α × β × √3 × (−√3) = \(\frac{e}{a}\)

= \(\frac{-3}{2}\) 

⇒ −3αβ = \(\frac{-3}{2}\)

⇒ αβ = \(\frac{1}{2}\)

Now,

(α-β)² = (α+β)² − 4αβ

= (\(\frac{3}{2}\))² - 4 × \(\frac{1}{2}\) 

= \(\frac{9}{4}\) - 2

= \(\frac{9-8}{4}\)

= \(\frac{1}{4}\).

∴ α-β = \(\frac{1}{2}\). … (2)

By, adding equations (1) and (2), we get

(α+β) + (α-β

= \(\frac{3}{2}\) + \(\frac{1}{2}\)

= \(\frac{4}{2}\) 

= 2

⇒ 2α = 2 

⇒ α = 1.

Now, 

Putting α = 1 in equation (1), we get

1 + β = \(\frac{3}{2}\)

⇒ β = \(\frac{3}{2}\) - 1

= \(\frac{1}{2}\).

Hence,

Two other zeros of given polynomial are 1 & \(\frac{1}{2}\).