What is (x-2)^2+(x-3)^2<= 2(x-1)^2 ?

The solution to (x-2)^2+(x-3)^2 = 11/6

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(x-2)^2+(x-3)^2<= 2(x-1)^2

(x - 2)^{2} + (x - 3)^{2} \leq 2 (x - 1)^{2}

x \geq \frac{11}{6}

Interval Notation

[\frac{11}{6}, \infty)

Decimal Notation

x \geq 1.83333 \dots

Solution steps

(x - 2)^{2} + (x - 3)^{2} \leq 2 (x - 1)^{2}

Rewrite in standard form

- 6 x + 11 \leq 0

Move

11

to the right side

- 6 x \leq - 11

Multiply both sides by

- 1

6 x \geq 11

Divide both sides by

6

x \geq \frac{11}{6}

22 - 12 k = - 14

30 n^{5} - 24 n^{3} - 12 n^{2}

\frac{4 ^{7 x - 10}}{4 ^{4 x + 32}}

2 a^{2} + 8 a = - 12

- 3 ∣ x ∣ = 0

What is (x-2)^2+(x-3)^2<= 2(x-1)^2 ?
The solution to (x-2)^2+(x-3)^2<= 2(x-1)^2 is x>= 11/6