Question

If α,β and $\gamma$ are zeroes of cubic polynomial kt³-5t+9, also α³+β³+$\gamma^3$=27, find the value of 'k'.

Answer 1

Let the roots be p , q , r for my convenience .

k t³ - 5 t + 9 = 0

==> k t³ + 0 t² - 5 t + 9 = 0

Comparing with  a x³ + bx² + c x+ d we get :

a = k

b = 0

c = - 5

d = 9

Sum of roots

We know that sum of roots = -b / a

==> p + q + r = -0/k = 0

We know that if p + q + r = 0

then : p³ + q³ + r³ = 3 pqr

Given : p³ + q³ + r³ = 27

3 pqr = 27

==> pqr = 27/3

             = 9 ........................(1)

Product of roots

We know that product of roots = -d/a

                                                   = -9/k

Hence pqr =  - 9/k

But see (1) ==> 9 = -9/k

                ==> k = -9/9

                ==> k = -1

The value of k is  -1

Hope it helps u :-)

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Answer 2

Answer:

k = -1

Step-by-step explanation:

Given cubic polynomial is kt³ - 5t + 9.

Here, a = k, b = 0, c = -5, d = 9.

Given that α,β,γ are zeroes of  polynomial.

Calculate roots:

α + β + γ = 0

αβ + βγ + γα = 5/k.

αβγ = -9/k

Now,

We know that if α + β + γ = 0, then α³ + β³ + γ³ = 3αβγ

⇒ 27 = 3(-9/k)

⇒ 27 = -27/k

⇒ 27k = -27

⇒ k = -1.

Therefore, the value of k = -1.

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