0.6 is greater than 0.5 or less please help

Answer:

4

Step-by-step explanation: the area is width x length so 2 x 2

The possible values of x that satisfy the equation are

x = ±√3 and x = ±2√3.

Here, we have,

To find a value of x that is a solution to the equation, we need to solve the equation for x.

Let's go through the steps:

Start with the given equation:

(x² - 8)² + x² - 8 = 20

Expand the squared term (x² - 8)²:

(x² - 8)(x² - 8) + x² - 8 = 20

Simplify:

(x⁴ - 16x² + 64) + x² - 8 = 20

Combine like terms:

x⁴ - 16x² + x² + 64 - 8 = 20

Simplify:

x⁴ - 15x² + 56 = 20

Rearrange the equation to bring all terms to one side:

x⁴ - 15x²+ 56 - 20 = 0

Simplify:

x⁴ - 15x² + 36 = 0

Now, we have a quadratic equation in terms of x².

Let's substitute y = x² to simplify further:

y² - 15y + 36 = 0

Solve the quadratic equation for y:

(y - 3)(y - 12) = 0

This gives us two possible values for y:

y - 3 = 0 --> y = 3

y - 12 = 0 --> y = 12

Substitute back y = x²:

For y = 3:

x² = 3 --> x = ±√3

For y = 12:

x² = 12 --> x = ±√12 = ±2√3

So, the possible values of x that satisfy the equation are x = ±√3 and x = ±2√3.

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Answer:

x = 2 sqrt(3) or x = -2 sqrt(3) or x = sqrt(3) or x = -sqrt(3)

Step-by-step explanation:

Solve for x:

-8 + x^2 + (x^2 - 8)^2 = 20

Expand out terms of the left hand side:

x^4 - 15 x^2 + 56 = 20

Subtract 20 from both sides:

x^4 - 15 x^2 + 36 = 0

Substitute y = x^2:

y^2 - 15 y + 36 = 0

The left hand side factors into a product with two terms:

(y - 12) (y - 3) = 0

Split into two equations:

y - 12 = 0 or y - 3 = 0

Add 12 to both sides:

y = 12 or y - 3 = 0

Substitute back for y = x^2:

x^2 = 12 or y - 3 = 0

Take the square root of both sides:

x = 2 sqrt(3) or x = -2 sqrt(3) or y - 3 = 0

Add 3 to both sides:

x = 2 sqrt(3) or x = -2 sqrt(3) or y = 3

Substitute back for y = x^2:

x = 2 sqrt(3) or x = -2 sqrt(3) or x^2 = 3

Take the square root of both sides:

Answer: x = 2 sqrt(3) or x = -2 sqrt(3) or x = sqrt(3) or x = -sqrt(3)