Which of the following is not a quadratic equation?
\(2(x-1)^2=4x^2-2x+1\)
\(2x-x^2=x^2+5\)
\((\sqrt{2}x+\sqrt{3})^2+x^2=3x^2-5x\)
\((x^2+2x)^2=x^4+3+4x^3\)
Step 1: Recall what a quadratic equation is.
A quadratic equation is any equation that can be written in the form \(ax^2 + bx + c = 0\), where:
Step 2: Check each option.
(A) \(2(x-1)^2 = 4x^2 - 2x + 1\)
Expand left side: \(2(x^2 - 2x + 1) = 2x^2 - 4x + 2\).
Equation: \(2x^2 - 4x + 2 = 4x^2 - 2x + 1\).
Bring all terms to one side: \(-2x^2 - 2x + 1 = 0\) or \(2x^2 + 2x - 1 = 0\).
This is quadratic (highest power is 2).
(B) \(2x - x^2 = x^2 + 5\)
Rearrange: \(-x^2 + 2x = x^2 + 5\).
Bring terms together: \(-2x^2 + 2x - 5 = 0\) or \(2x^2 - 2x + 5 = 0\).
This is also quadratic (highest power is 2).
(C) \((\sqrt{2}x + \sqrt{3})^2 + x^2 = 3x^2 - 5x\)
Expand: \(2x^2 + 2\sqrt{6}x + 3 + x^2 = 3x^2 - 5x\).
So LHS = \(3x^2 + 2\sqrt{6}x + 3\).
Equation: \(3x^2 + 2\sqrt{6}x + 3 = 3x^2 - 5x\).
Cancel \(3x^2\) on both sides → \(2\sqrt{6}x + 3 = -5x\).
Combine terms: \((2\sqrt{6} + 5)x + 3 = 0\).
This is linear (highest power is 1).
(D) \((x^2 + 2x)^2 = x^4 + 3 + 4x^3\)
Expand LHS: \(x^4 + 4x^3 + 4x^2\).
Equation: \(x^4 + 4x^3 + 4x^2 = x^4 + 4x^3 + 3\).
Cancel \(x^4\) and \(4x^3\) on both sides → \(4x^2 = 3\).
This is quadratic (highest power is 2).
Final Answer: Option (C) is not quadratic because after simplification, it becomes a linear equation.