2, 2x2 + x + 1 = 0

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Solution

We have 2 x 2 +x+1=0 .

Any equation of the form a x 2 +bx+c=0 has the solution

x= −b± D 2a

Here, D= b 2 −4ac is called the discriminant of the quadratic equation.

Therefore,

x= −1± 1 2 −4×2×1 2×2 = −1± −7 4 = −1± 7 i 4

Thus, x= −1± 7 i 4

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6x^2-x-2=0 , x=-1/2,2/3

∣ ∣ ∣ \frac{| x | + 2}{| x^{2} | + 2 | x |} ∣ ∣ ∣ > \frac{1}{2}

x \in

f (x) = \{\begin{cases} - 1, & for - 2 \leq x \leq 0 \\ x - 1, & for 0 \leq x \leq 2 \end{cases}

\{- \frac{1}{2}\}

f : [- 2, 2] \to R

f (x) = {\begin{matrix} - 1, - 2 \leq x < 0 x - 1, 0 \leq x \leq 2 \end{matrix}

{x | x \in [- 2, 2] : x \leq 0 and f (| x |) = x}

f (x) = \{\begin{matrix} x^{2} / a, & 0 \leq x < 1 \\ a, & 1 \leq x < \sqrt{2} \\ \frac{2 b^{2} - 4 b}{x^{2}}, & \sqrt{2} \leq x < \infty \end{matrix}

\sqrt{2}

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