Question:
If one root of the cubic equation
\[
x^3+36=7x^2
\]
is double of another, then the number of negative roots is
If one root of the cubic equation
\[
x^3+36=7x^2
\]
is double of another, then the number of negative roots is
Show 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.
Updated On: Jun 22, 2026
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The Correct Option is A
Solution and Explanation
Step 1: Write the equation in standard form.
Given, \[ x^3+36=7x^2 \] So, \[ x^3-7x^2+36=0 \]
Step 2: Try simple integral roots.
Substitute \[ x=3 \] \[ 3^3-7(3)^2+36=27-63+36=0 \] So, \[ x=3 \] is a root.
Step 3: Factorize the polynomial.
Since \(x=3\) is a root, \[ x-3 \] is a factor.
Now, \[ x^3-7x^2+36=(x-3)(x^2-4x-12) \] \[ =(x-3)(x-6)(x+2) \]
Step 4: Identify all roots.
Thus, the roots are \[ 3,\quad 6,\quad -2 \] Here, \[ 6=2\times 3 \] So, one root is double of another.
Step 5: Count the negative roots.
Among \[ 3,\quad 6,\quad -2, \] only \[ -2 \] is negative.
Step 6: Final conclusion.
Therefore, the number of negative roots is \[ \boxed{1} \]
Given, \[ x^3+36=7x^2 \] So, \[ x^3-7x^2+36=0 \]
Step 2: Try simple integral roots.
Substitute \[ x=3 \] \[ 3^3-7(3)^2+36=27-63+36=0 \] So, \[ x=3 \] is a root.
Step 3: Factorize the polynomial.
Since \(x=3\) is a root, \[ x-3 \] is a factor.
Now, \[ x^3-7x^2+36=(x-3)(x^2-4x-12) \] \[ =(x-3)(x-6)(x+2) \]
Step 4: Identify all roots.
Thus, the roots are \[ 3,\quad 6,\quad -2 \] Here, \[ 6=2\times 3 \] So, one root is double of another.
Step 5: Count the negative roots.
Among \[ 3,\quad 6,\quad -2, \] only \[ -2 \] is negative.
Step 6: Final conclusion.
Therefore, the number of negative roots is \[ \boxed{1} \]
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