The solution to x^2<= 4x+1 is -sqrt(5)+2<= x<= sqrt(5)+2

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x^2<= 4x+1

The solution to x^2<= 4x+1 is -sqrt(5)+2<= x<= sqrt(5)+2

Solution

x^{2} \leq 4 x + 1

- 5 + 2 \leq x \leq 5 + 2

Interval Notation

[- 5 + 2, 5 + 2]

Decimal Notation

- 0.23606 \dots \leq x \leq 4.23606 \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} - 5

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

Move

5

to the right side

(x - 2)^{2} \leq 5

For

u^{n} \leq a

, if

n

is even then

- 5 \leq x - 2 \leq 5

a \leq u \leq b

then

a \leq u and u \leq b

- 5 \leq x - 2 and x - 2 \leq 5

- 5 \leq x - 2 : x \geq - 5 + 2

x - 2 \leq 5 : x \leq 5 + 2

Combine the intervals

x \geq - 5 + 2 and x \leq 5 + 2

Merge Overlapping Intervals

- 5 + 2 \leq x \leq 5 + 2

x^{2} - 4 < - 2 x + 4

d^{2} + 5 d - 24

4.2 < - 1.9 + x

5 y + 9 \leq 4

- 2 ∣ 5 - 5 x ∣ - 10 = - 26