If k and 2 k are zeros of f x - x 3-4 x 2-9 kx -90- find k and all three zeros of f x -A- K-1- roots -3-6-5B- K-3- roots -3-6-5C- K -3- roots -3-6-5D- K-3- roots -3-6-5

The correct option is B K = 3, roots = 3, 6, 5Since k and 2k are zeros of f(x) f(k) = 0 ⇒ k^3+4k^2+9k^2 90=0 ⇒ k^3+13k^2 90=0 ....(i) And f(2k) = 0 ⇒ 8k^3+ ...

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Standard IX

Mathematics

Zeroes of a Polynomial

If k and 2 k ...

Question

If k and 2k are zeros of $f (x) = x^{3} + 4 x^{2} + 9 k x = 90$ , find k and all three zeros of f(x).

K = 1, roots = -3, -6, -5

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K = -3, roots = -3, -6, 5

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K = -3, roots = -3, 6, -5

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K = 3, roots = -3, -6, 5

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Solution

The correct option is B K = -3, roots = -3, -6, 5
Since k and 2k are zeros of f(x)
f(k) = 0
$\Rightarrow k^{3} + 4 k^{2} + 9 k^{2} - 90 = 0 \Rightarrow k^{3} + 13 k^{2} - 90 = 0 . . . . (i)$
And f(2k) = 0
$\Rightarrow 8 k^{3} + 16 k^{2} + 18 k^{2} - 90 = 0 \Rightarrow 8 k^{3} + 34 k^{2} - 90 = 0 \Rightarrow 4 k^{3} + 17 k^{2} - 45 = 0 . . . . (i i)$
Multiplying equation (i) by 4 and then subtracting from (ii), we get

$- 35 k^{2} + 315 = 0 \Rightarrow k^{2} = 9 \Rightarrow k = \pm 3$
k = 3 does not satisfy the given polynomial.
$∴ k = - 3$
The two roots are – 3 and – 6
Sum of roots = - 4
$\Rightarrow - 3 - 6$ +third root =-4
$\Rightarrow$ third root =5

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