The solution to x^2<= 6x-4 is -sqrt(5)+3<= x<= sqrt(5)+3

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x^2<= 6x-4

The solution to x^2<= 6x-4 is -sqrt(5)+3<= x<= sqrt(5)+3

Solution

x^{2} \leq 6 x - 4

- 5 + 3 \leq x \leq 5 + 3

Interval Notation

[- 5 + 3, 5 + 3]

Decimal Notation

0.76393 \dots \leq x \leq 5.23606 \dots

Solution steps

x^{2} \leq 6 x - 4

Rewrite in standard form

x^{2} + 4 - 6 x \leq 0

Complete the square

x^{2} + 4 - 6 x : (x - 3)^{2} - 5

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

Move

5

to the right side

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

For

u^{n} \leq a

, if

n

is even then

- 5 \leq x - 3 \leq 5

a \leq u \leq b

then

a \leq u and u \leq b

- 5 \leq x - 3 and x - 3 \leq 5

- 5 \leq x - 3 : x \geq - 5 + 3

x - 3 \leq 5 : x \leq 5 + 3

Combine the intervals

x \geq - 5 + 3 and x \leq 5 + 3

Merge Overlapping Intervals

- 5 + 3 \leq x \leq 5 + 3

(4)^{2}

ln (x^{y})

2 x^{2} + 54 = 0

6 x^{2} + 23 x + 15 = 0

(\frac{1}{6})^{2}