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Which of the following is not a quadratic equation:

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2 (x + 1)^{2} = 4 x^{2} + 2 x + 1

2 x + x^{2} = 2 x^{2} + 5

(\sqrt{2} \times \sqrt{3}

x)^{2} + x^{2} = 3 x^{2} + 5 x

(

x^{2} + 2 x) = x^{5} + 3 + 4 x^{3}

answer is D.

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Detailed Solution

For option A,

2 (x + 1)^{2} = 4 x^{2} + 2 x + 1

Using the formula,

(a + b)^{2} = a^{2} + 2 ab + b^{2}

We get,

2 (x + 1) = 4 x^{2} + 2 x + 1

2 (x^{2} + 2 x + 1) = 4 x^{2} + 2 x + 1

2 x^{2} + 4 x + 2 = 4 x^{2} + 2 x + 1

2 x^{2} - 2 x - 1 = 0

Here the highest power of the variable is 2, hence it is a quadratic equation.
For option B,

2 x + x^{2} = 2 x^{2} + 5

2 x + x^{2} = 2 x^{2} + 5

x^{2} - 2 x + 5 = 0

The degree 2 shows that it is a quadratic equation.
For option C,

(\sqrt{2} \times \sqrt{3}

x)^{2} + x^{2} = 3 x^{2} + 5 x

(

\sqrt{6} x)^{2} + x^{2} = 3 x^{2} + 5 x

6 x^{2} + x^{2} = 3 x^{2} + 5 x

7 x^{2} = 3 x^{2} + 5 x

4 x^{2} - {5 x}^{} = 0

Here highest power is 2, so this is also the wrong option.
For option D, (

x^{2} + 2 x) = x^{5} + 3 + 4 x^{3}

The degree of the equation is 5, thus it is not a quadratic equation.

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