Question

If a, b are zeroes of polynomial p(x) =5x²+5x+1 then find the value of (1) a²+b² (ii) a-¹+b‐¹​

Answer 1

Answer

i) 3/5

ii) -5

$\rule{200}1$

Explanation

Given,

zeroes of polynomial p(x) = 5x² + 5x + 1 are a, b

To find,

the value of a² + b² and 1/a + 1/b

We know that sum of zeroes = $\sf -\frac{coefficient\ of\ x}{coefficient\ of\ x^2}$

So, we get a + b = -5/5 = -1

And, product of zeroes = $\sf \frac{constant}{coefficient\ of\ x^2}$

→ ab = 1/5

Now,

i) a² + b² = (a + b)² - 2ab

→ a² + b² = (-1)² - 2/5

→ a² + b² = 1 - 2/5

→ a² + b² = 3/5

ii) 1/a + 1/b = (a + b)/ab

→ $\sf\frac{-1}{\frac{1}{5}}$

→ 1/a + 1/b = -5

Answer 2

$\huge\mathfrak\green{Heyaa!!}$

$\huge\mathfrak\red{Answer:-}$

The given polynomial is

p (x) = 5x² + 5x + 1

And Since α and β are the zeroes of p (x),

α + β = - 5/5

⇒ α + β = - 1 ...(1)

and

αβ = 1/5 ...(2)

i) Now, α² + β²

= (α + β)² - 2αβ

= (- 1)² - 2 (1/5)

= 1 - 2/5

= (5 - 2)/5

= 3/5

(ii) Now, α^(- 1) + β^(- 1)

= 1/α + 1/β

= (β + α)/(αβ)

= (- 1)/(1/5)

= - 5