What is x^2<= 8x-11 ?

The solution to x^2<= 8x-11 is -sqrt(5)+4<= x<= sqrt(5)+4

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x^2<= 8x-11

x^{2} \leq 8 x - 11

- 5 + 4 \leq x \leq 5 + 4

Solution steps

x^{2} \leq 8 x - 11

Rewrite in standard form

x^{2} + 11 - 8 x \leq 0

Complete the square

x^{2} + 11 - 8 x : (x - 4)^{2} - 5

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

Move

5

to the right side

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

For

u^{n} \leq a

, if

n

is even then

- 5 \leq x - 4 \leq 5

a \leq u \leq b

then

a \leq u and u \leq b

- 5 \leq x - 4 and x - 4 \leq 5

- 5 \leq x - 4 : x \geq - 5 + 4

x - 4 \leq 5 : x \leq 5 + 4

Combine the intervals

x \geq - 5 + 4 and x \leq 5 + 4

Merge Overlapping Intervals

- 5 + 4 \leq x \leq 5 + 4

- 6 x - 7 = 5 x - 117

30 x^{2} - 50 x

49 + 9

(9 m n^{6})^{\frac{1}{2}}

x^{2} - 18 x + 65 = 0