AP EAMCET · Maths · Quadratic Equation
If \(\mathrm{a}\) and \(\mathrm{b}\) are the roots of the equation \(\mathrm{y}^2+\mathrm{y}+1=0\), then the value of \(a^4+b^4+a^{-1} b^{-1}\) is
- A \(1\)
- B \(0\)
- C \(5\)
- D \(2\)
Answer & Solution
Correct Answer
(B) \(0\)
Step-by-step Solution
Detailed explanation
Given the equation \(y^2+y+1=0\) \(\begin{aligned} & \text { Now, } a^4+b^4+a^{-1} b^{-1}=a^4+b^4+\frac{1}{a b} \\ & =\left(a^2+b^2\right)^2-2 a^2 b^2+\frac{1}{1} \\ & =\left((a+b)^2-2 a b\right)^2-2 \times 1+1 \\ & =\left((-1)^2-2\right)^2-1=1-1=0 \end{aligned}\)
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