Question

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

• A. \(1\)
• B. \(2\)
• C. \(3\)
• D. \(0\)

Hint:

For cubic equations, first bring the equation into standard form and then try simple integral roots. After factorization, count the roots according to the condition asked in the question.

Answer 1

The Correct Option is A

Step 1: Rewrite the equation in standard form.
The equation $x^3 + 36 = 7x^2$ becomes $x^3 - 7x^2 + 36 = 0$.
Step 2: Use the condition that one root is double another.
Let the roots be $\alpha$, $2\alpha$, and $\beta$. By Vieta's formulas for $x^3 - 7x^2 + 0 \cdot x + 36 = 0$:
Sum: $\alpha + 2\alpha + \beta = 7 \Rightarrow 3\alpha + \beta = 7$.
Product of roots: $\alpha \cdot 2\alpha \cdot \beta = -36 \Rightarrow 2\alpha^2 \beta = -36$.
Sum of products in pairs: $2\alpha^2 + \alpha\beta + 2\alpha\beta = 0 \Rightarrow 2\alpha^2 + 3\alpha\beta = 0$.
Step 3: Solve the system for $\alpha$.
From $2\alpha^2 + 3\alpha\beta = 0$: either $\alpha=0$ (trivial, rejected since product $= -36 \neq 0$) or $2\alpha + 3\beta = 0 \Rightarrow \beta = -\frac{2\alpha}{3}$.
Substitute into $3\alpha + \beta = 7$: $3\alpha - \frac{2\alpha}{3} = 7 \Rightarrow \frac{7\alpha}{3} = 7 \Rightarrow \alpha = 3$. Then $\beta = -2$.
Step 4: Verify the roots are $3, 6, -2$.
Product: $3 \times 6 \times (-2) = -36$ ✓. Sum: $3+6-2=7$ ✓. Check $x=-2$: $-8-7(4)+36 = -8-28+36=0$ ✓.
Step 5: Identify negative roots.
The roots are $3, 6, -2$. Only $-2$ is negative. So the number of negative roots is $1$.
Step 6: State the answer.
Among all roots $3, 6, -2$, exactly one root is negative. \[ \boxed{1} \]