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AP EAMCET · Maths · Quadratic Equation

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

  1. A \(1\)
  2. B \(2\)
  3. C \(3\)
  4. D \(0\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(1\)

Step-by-step Solution

Detailed explanation

Let \(a, 2 a\) and \(b\) be roots of the cubic equation \(x^3+36-7 x^2=0\) So, sum of roots \(=-\frac{(-7)}{1}\) \(\therefore \quad a+2 a+b=7\) \(\Rightarrow \quad 3 a+b=7\) ...(i) Now, \(a(2 a) b=36\) \(\left[\right.\) product of roots \(\left.=\frac{d}{a}\right]\)…