(x^2-7)^2+2x^2-14=0

2 min read Jun 17, 2024

Solving the Equation (x^2-7)^2 + 2x^2 - 14 = 0

This equation might look intimidating at first, but we can solve it systematically using algebraic manipulation and substitution. Let's break down the process:

1. Substitution

To simplify the equation, let's introduce a new variable:

Let y = x² - 7

Substituting this into the original equation, we get:

y² + 2y - 14 = 0

Now we have a quadratic equation in terms of 'y', which is much easier to handle.

2. Solving the Quadratic Equation

We can solve this quadratic equation using the quadratic formula:

y = (-b ± √(b² - 4ac)) / 2a

Where a = 1, b = 2, and c = -14.

Plugging in the values:

y = (-2 ± √(2² - 4 * 1 * -14)) / 2 * 1

y = (-2 ± √(60)) / 2

y = (-2 ± 2√15) / 2

y = -1 ± √15

Therefore, we have two possible solutions for 'y':

y₁ = -1 + √15 y₂ = -1 - √15

3. Solving for x

Now, we need to substitute back the original expression for 'y' and solve for 'x':

For y₁ = -1 + √15:

x² - 7 = -1 + √15

x² = 6 + √15

x = ±√(6 + √15)

For y₂ = -1 - √15:

x² - 7 = -1 - √15

x² = 6 - √15

x = ±√(6 - √15)

4. Final Solutions

Therefore, the solutions for the original equation (x² - 7)² + 2x² - 14 = 0 are:

x₁ = √(6 + √15)
x₂ = -√(6 + √15)
x₃ = √(6 - √15)
x₄ = -√(6 - √15)

These are the four distinct solutions to the equation.