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For the equation, $x^{3} - {7 x}^{2} + 36 = 0$ if one root is twice the other, then find the other root.

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-2

All of the above

answer is D.

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Detailed Solution

Let the roots of the equation are a,2a,b then
The sum of the roots
a+2a+b = 7…..(i)
The product of the roots

a \times 2 a \times b

= -36
Or,

{2 a}^{2} b = - 36

……(ii)
And

{2 a}^{2}

+ ab + 2ab = 0
or, a(2a+3b) = 0
so, b =

\frac{- 2 a}{3}

…..(iii)
From (ii) we get,

\frac{- 4 a^{3}}{3} = - 36

Or,

a^{3} = \frac{108}{4} = 27

Or, a = 3
Then, 2a =

3 \times 2 = 6

From (iii)
b=

\frac{- 6}{3} = - 2

Hence the roots are 3, 6 and -2.
So, all of the above are correct.

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