TS EAMCET · Maths · Quadratic Equation
If each root of the equation \(2 x^3+a x^2-8 x+b=0\) is reduced by one, then in the transformed equation thus formed, the term containing \(x^2\) and the constant term are vanishing. The roots of the original equation are
- A \(1,-3,2\)
- B \(1,1 \pm \sqrt{7}\)
- C \(1,1,-6\)
- D \(1,3 \sqrt{2},-\sqrt{2}\)
Answer & Solution
Correct Answer
(B) \(1,1 \pm \sqrt{7}\)
Step-by-step Solution
Detailed explanation
Let, the roots of equation \(2 x^3+a x^2-8 x+b=0\) are \(\alpha, \beta, \gamma\) After reducing by one from the roots, the quantities becomes \(\alpha-1, \beta-1, \gamma-1\). Now, let \(x=\alpha-1\) \(\Rightarrow \quad \alpha=x+1\) Since, \(\alpha\) is the root of given…
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