The solution to x^2<= 4x-1 is -sqrt(3)+2<= x<= sqrt(3)+2

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x^2<= 4x-1

The solution to x^2<= 4x-1 is -sqrt(3)+2<= x<= sqrt(3)+2

Solution

x^{2} \leq 4 x - 1

- 3 + 2 \leq x \leq 3 + 2

Interval Notation

[- 3 + 2, 3 + 2]

Decimal Notation

0.26794 \dots \leq x \leq 3.73205 \dots

Solution steps

x^{2} \leq 4 x - 1

Rewrite in standard form

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

Complete the square

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

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

Move

3

to the right side

(x - 2)^{2} \leq 3

For

u^{n} \leq a

, if

n

is even then

- 3 \leq x - 2 \leq 3

a \leq u \leq b

then

a \leq u and u \leq b

- 3 \leq x - 2 and x - 2 \leq 3

- 3 \leq x - 2 : x \geq - 3 + 2

x - 2 \leq 3 : x \leq 3 + 2

Combine the intervals

x \geq - 3 + 2 and x \leq 3 + 2

Merge Overlapping Intervals

- 3 + 2 \leq x \leq 3 + 2

lo g_{15} (15)

\frac{y ^{- 2}}{y ^{2}}

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