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

The solution to (x-1)(x+1)>= 2(x^2-3) is -sqrt(5)<= x<= sqrt(5)

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

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

- 5 \leq x \leq 5

Interval Notation

[- 5, 5]

Decimal Notation

- 2.23606 \dots \leq x \leq 2.23606 \dots

Solution steps

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

Rewrite in standard form

- x^{2} + 5 \geq 0

Move

5

to the right side

- x^{2} \geq - 5

Multiply both sides by

- 1

x^{2} \leq 5

For

u^{n} \leq a

, if

n

is even then

- 5 \leq x \leq 5

16 t^{2} - 52 t - 40

\frac{1}{4} x^{2} = 5 - x^{2}

x^{2} + 3 x - 10 < 0

\frac{x}{4} + 6 = 8

y - 1 = 0

What is (x-1)(x+1)>= 2(x^2-3) ?
The solution to (x-1)(x+1)>= 2(x^2-3) is -sqrt(5)<= x<= sqrt(5)