If $4 x^4=5 x^2-1, x>\frac{1}{\sqrt{2}}$, then what is the value of $\left(2 x^2-x-1\right)$ ?
Answer & explanation
Correct answer: option 3
4x4 - 5x2 + 1 = 0
= 4x4 - 4x2 - x2 + 1 = 0
= 4x2(x2 - 1) – 1(x2 - 1) = 0
= (x2 - 1)(4x2 – 1) = 0
= (x2 - 1) = 0 (4x2 – 1) = 0
x - 1 = 0
2x – 1 = 0
x = 1
another value of x = \(\frac{1}{2}\)
(2x2 - x - 1)
Put x = 1
2 × 1 - 1 - 1
= 2 - 2 = 0