What is 0>= 2x^2-4x+1 ?

The solution to 0>= 2x^2-4x+1 is -sqrt(1/2)+1<= x<= sqrt(1/2)+1

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0>= 2x^2-4x+1

0 \geq 2 x^{2} - 4 x + 1

- \frac{1}{2} + 1 \leq x \leq \frac{1}{2} + 1

Solution steps

0 \geq 2 x^{2} - 4 x + 1

Switch sides

2 x^{2} - 4 x + 1 \leq 0

Complete the square

2 x^{2} - 4 x + 1 : 2 (x - 1)^{2} - 1

2 (x - 1)^{2} - 1 \leq 0

Move

1

to the right side

2 (x - 1)^{2} \leq 1

Divide both sides by

2

(x - 1)^{2} \leq \frac{1}{2}

For

u^{n} \leq a

, if

n

is even then

- \frac{1}{2} \leq x - 1 \leq \frac{1}{2}

a \leq u \leq b

then

a \leq u and u \leq b

- \frac{1}{2} \leq x - 1 and x - 1 \leq \frac{1}{2}

- \frac{1}{2} \leq x - 1 : x \geq - \frac{1}{2} + 1

x - 1 \leq \frac{1}{2} : x \leq \frac{1}{2} + 1

Combine the intervals

x \geq - \frac{1}{2} + 1 and x \leq \frac{1}{2} + 1

Merge Overlapping Intervals

- \frac{1}{2} + 1 \leq x \leq \frac{1}{2} + 1

\frac{x ^{- \frac{1}{3}}}{x ^{\frac{2}{3}}}

\frac{20 !}{4 ! 16 !}

3 x^{5} - 14 x^{4} - 14 x^{3} + 36 x^{2} + 43 x + 10

10 x^{2} - 33 x - 7

x^{2} + 13 x y - 68 y^{2}

What is 0>= 2x^2-4x+1 ?
The solution to 0>= 2x^2-4x+1 is -sqrt(1/2)+1<= x<= sqrt(1/2)+1