What is x^2-3x+1<= 0 ?

The solution to x^2-3x+1<= 0 is (-sqrt(5)+3)/2 <= x<= (sqrt(5)+3)/2

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x^2-3x+1<= 0

x^{2} - 3 x + 1 \leq 0

\frac{- 5 + 3}{2} \leq x \leq \frac{5 + 3}{2}

Solution steps

x^{2} - 3 x + 1 \leq 0

Complete the square

x^{2} - 3 x + 1 : (x - \frac{3}{2})^{2} - \frac{5}{4}

(x - \frac{3}{2})^{2} - \frac{5}{4} \leq 0

Move

\frac{5}{4}

to the right side

(x - \frac{3}{2})^{2} \leq \frac{5}{4}

For

u^{n} \leq a

, if

n

is even then

- \frac{5}{4} \leq x - \frac{3}{2} \leq \frac{5}{4}

a \leq u \leq b

then

a \leq u and u \leq b

- \frac{5}{4} \leq x - \frac{3}{2} and x - \frac{3}{2} \leq \frac{5}{4}

- \frac{5}{4} \leq x - \frac{3}{2} : x \geq \frac{- 5 + 3}{2}

x - \frac{3}{2} \leq \frac{5}{4} : x \leq \frac{5 + 3}{2}

Combine the intervals

x \geq \frac{- 5 + 3}{2} and x \leq \frac{5 + 3}{2}

Merge Overlapping Intervals

\frac{- 5 + 3}{2} \leq x \leq \frac{5 + 3}{2}

2 lo g_{7} (- 2 r)

4 x^{2} + 11 x - 20

z^{2} + 2 z + 1 - w^{2}

\frac{5 n ^{2} - 245}{7 n ^{2} + 112 n + 441}

\frac{3}{4} - \frac{7}{8}

What is x^2-3x+1<= 0 ?
The solution to x^2-3x+1<= 0 is (-sqrt(5)+3)/2 <= x<= (sqrt(5)+3)/2