TS EAMCET · Maths · Quadratic Equation
The maximum possible number of real roots of the equation \(x^{\frac{5}{5}}-6 x^2-4 x+5=0\) is
- A \(0\)
- B \(3\)
- C \(4\)
- D \(5\)
Answer & Solution
Correct Answer
(B) \(3\)
Step-by-step Solution
Detailed explanation
Let \(f(x)=x^2-6 x^2-4 x+5\) \[ \Rightarrow f(-x)=-x^5-6 x^2+4 x+5 \] Number of changes of sign in \(f(x)\) are 2 and number of changes of sign in \(f(-x)\) are 1 . \(\therefore\) By descarte's rule of signs Maximum number of +ve real roots are 2 and \(-\mathrm{ve}\) real roots…
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