AP EAMCET · Maths · Quadratic Equation
Roots of the equation \(a(b-c) x^2+b(c-a) x+c(a-b)\) \(=0\) are
- A \(\frac{a(b-c)}{c(a-b)}, 1\)
- B \(\frac{b(c-a)}{c(a-b)}, 1\)
- C \(\frac{c(a-b)}{a(b-c)}, 1\)
- D \(\frac{c(a-b)}{b(c-a)}, 1\)
Answer & Solution
Correct Answer
(C) \(\frac{c(a-b)}{a(b-c)}, 1\)
Step-by-step Solution
Detailed explanation
\(a(b-c) x^2+b(c-a) x+c(a-b)=0\) \(x=1\) satisfies his equation Product of roots \(=\frac{c(a-b)}{a(b-c)}=1 \times \alpha\) \(\alpha=\frac{c(a-b)}{a(b-c)}=\text { other root }\) Roots are \(\frac{c(a-b)}{a(b-c)}\) and 1
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