AP EAMCET · Maths · Quadratic Equation
Suppose that the equation \(a x^2+b x+c=0\) has roots \(\alpha\) and \(\beta\), both of which are different from \(\frac{1}{3}\), then an equation whose roots are \(\frac{1}{3 \alpha-1}\) and \(\frac{1}{3 \beta-1}\) is
- A \((a+3 b+9 c) x^2+(3 b+2 a) x+a=0\)
- B \((a+3 b+9 c) x^2-(3 b+2 a) x+a=0\)
- C \((a+3 b+9 c) x^2+(3 b-2 a) x+a=0\)
- D \((a+3 b+9 c) x^2-(3 b-2 a) x+a=0\)
Answer & Solution
Correct Answer
(A) \((a+3 b+9 c) x^2+(3 b+2 a) x+a=0\)
Step-by-step Solution
Detailed explanation
\(a x^2+b x+c=0...(i)\) Let \(\alpha\) and \(\beta\) are the roots of Eq. (i). \[ \alpha+\beta=-\frac{b}{a} \text { and } \alpha \beta=\frac{c}{a} \] Now, the quadratic equation whose roots are…
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