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

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Aicharya · Accepted Answer

Let $a, 2 a$ and $b$ be roots of the cubic equation $x^3+36-7 x^2=0$ So, sum of roots $=-\frac{(-7)}{1}$ $	herefore \quad a+2 a+b=7$ $\Rightarrow \quad 3 a+b=7$ ...(i) Now, $a(2 a) b=36$ $\left[ight.$ product of roots $\left.=\frac{d}{a}ight]$ $2 a^2 b=36$ and $a b+a(2 a)+2 a(b)=0$ $\begin{aligned} \Rightarrow & & 2 a^2+3 a b & =0 \Rightarrow a(2 a+3 b)=0 \ \Rightarrow & & b & =\frac{-2 a}{3}\end{aligned}$ $\Rightarrow \quad 3 a-\frac{2 a}{3}=7 \quad$ [from Eq. (i)] $\Rightarrow \quad \frac{9 a-2 a}{3}=7 \Rightarrow 9 a-2 a=21$ $\Rightarrow \quad 7 a=21 \Rightarrow a=3$ and $b=7-3 a$ $=7-3(3)=-2$ [from Eq. (i)] Roots are $3,6,-2$ $	herefore$ Number of negative root is 1 .

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

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