If one root of the cubic equation x^3-7x^2+36 | vedclass

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(A) Let the roots of the cubic equation $x^3-7x^2+36=0$ be $a, 2a,$ and $b$.From the relation between roots and coefficients:Sum of roots: $a + 2a + b

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$1$

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(A) Let the roots of the cubic equation $x^3-7x^2+36=0$ be $a, 2a,$ and $b$.From the relation between roots and coefficients:Sum of roots: $a + 2a + b = 7 \Rightarrow 3a + b = 7$ ... $(i)$Sum of product of roots taken two at a time: $a(2a) + 2a(b) + b(a) = 0$ (since coefficient of $x$ is $0$)$2a^2 + 3ab = 0 \Rightarrow a(2a + 3b) = 0$Since $a$ cannot be $0$ (as $36 
eq 0$),we have $2a + 3b = 0 \Rightarrow b = -\frac{2a}{3}$.Substitute $b$ in $(i)$: $3a - \frac{2a}{3} = 7$ $\Rightarrow \frac{7a}{3} = 7$ $\Rightarrow a = 3$.Then $b = 7 - 3(3) = -2$.The roots are $a=3, 2a=6, b=-2$.The roots are $3, 6, -2$.Thus,there is only $1$ negative root.

If one root of the cubic equation $x^3-7x^2+36=0$ is double of another,then the number of negative roots is

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