Question

::If K and 2K are zeroes of f(x)=x^3+4x^2+9kx-90 ,find K and all three zeroes of f(x)
"please answer fast "​

Answer 1

Step-by-step explanation:

f(x) = x³ + 4x² + 9kx - 90

Then f(k)=k³ + 4k² +9k x k-90=0

⇒k³ + 4k² + 9k² - 90 = 0

⇒k³ + 13k² - 90=0

⇒k³ + 13k² = 90. (1)

And f(2k) = (2k)³ + 4(2k)² + 9(2k) x

k - 90 = 0

⇒ 8k³ + 16k² + 18k²-90=0

8k³ +34k290 = 0

⇒ 8k³ +34k² = 90. (2)

Equ. (2) multiply by 13 and (1) multiply by 17

Then 52k³ +221k² = 585

4k³ +221k² = 1530

solving (1) and (2) Then 35k³ = -945

$= > k {}^{3} = \frac{ - 945}{35}$

$= > k {}^{3} = - 27 = > k = - 3$

Then 2k = -6

$other \: root = \frac{ - (90)}{( - 3)( - 6)} = 5$

Then roots are -3, -6 and 5

$\huge{ son \: prodigal}$