# How can 2x^{2} = x^{2} + 4x + 8 be set up as a system of equations?

**Solution:**

Let y = 2x^{2} --------------- (1)

y = x^{2} + 4x + 8 --------------- (2)

By substituting the value of y in equation 2, we get

⇒ 2x^{2} = x^{2} + 4x + 8

⇒ 2x^{2} - x^{2} - 4x - 8 = 0

⇒ x^{2} - 4x = 8

By solving the equation using completing the square method, add 2^{2} to both sides of the above equation. We have

⇒ x^{2} - 4x + (2)^{2} = 8 + (2)^{2}

⇒ (x - 2)^{2 }= 8 + 4

By square root on both the sides we get

⇒ x - 2 = ± √12

⇒ x = 2 ± 2√3

From this we get two values of x

x = 2 + 2√3 or 2 - 2√3

The system of equations will have the same solutions for x.

Therefore, the system of quadratic equations can be set up as y = 2x^{2}, y = x^{2} + 4x + 8 and its solution is x = 2 + 2√3 or x = 2 - 2√3.

## How can 2x^{2} = x^{2} + 4x + 8 be set up as a system of equations?

**Summary:**

2x^{2} = x^{2} + 4x + 8 can be set up as a system of equations as y = 2x^{2}, y = x^{2} + 4x + 8 and its solution is x = 2 + 2√3 or x = 2 - 2√3.

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