TS EAMCET · Maths · Quadratic Equation
The number of all common roots of the equation \(x^4-10 x^3+37 x^2-60 x+36=0\) and the transformed equation of it obtained by increasing any two distinct roots of it by 1, keeping the other two roots fixed, is
- A \(1\)
- B \(3\)
- C \(4\)
- D \(2\)
Answer & Solution
Correct Answer
(B) \(3\)
Step-by-step Solution
Detailed explanation
\(P(x) = x^4-10 x^3+37 x^2-60 x+36\) \(P(x) = (x^2-5x+6)^2 = (x-2)^2(x-3)^2\) Roots of \(P(x)=0\) are \(\{2, 2, 3, 3\}\). Distinct roots are \(2\) and \(3\). Increase one \(2\) by \(1\) to get \(3\), and one \(3\) by \(1\) to get \(4\). The other two roots (one \(2\) and one…
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