# (2square root2-square root7)^x+(2square root2+square root 7)^x=<4square root2

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### 1 Answer

We notice that 2sqrt2 + sqrt7 = 1/(2sqrt2 - sqrt7).

We'll raise to x power both sides:

(2sqrt2 + sqrt7)^x = 1/(2sqrt2 - sqrt7)^x

We'll note (2sqrt2 + sqrt7)^x = a.

We'll re-write the inequality in a:

a + 1/a =< 4sqrt2

We'll multiply by a both sides:

a^2 + 1 =< 4asqrt2

We'll subtract 4asqrt2 both sides:

a^2 - 4asqrt2 + 1 =< 0

We'll find the roots of the equation a^2 - 4asqrt2 + 1 = 0

delta = 32 - 4 = 28

a1 = (4sqrt2 - 2sqrt7)/2

a1 = 2sqrt2 - sqrt7

a2 = 2sqrt2 + sqrt7

We'll come back to x variable:

(2sqrt2 + sqrt7)^x = a1

(2sqrt2 + sqrt7)^x = 2sqrt2 - sqrt7

(2sqrt2 + sqrt7)^x = 1/(2sqrt2 + sqrt7)

(2sqrt2 + sqrt7)^x = (2sqrt2 + sqrt7)^-1

Since the bases are matching, we'll get:

x = -1

(2sqrt2 + sqrt7)^x = a2

(2sqrt2 + sqrt7)^x = (2sqrt2 + sqrt7)

x = 1

**So, the inequality holds if x is in the range [-1 ; 1].**