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Which of the following is a quadratic equation?

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2 x^{2} - 5 x = {(x - 1)}^{2}

x^{3} - 3 \sqrt{x} + 2 = 0

x + \frac{1}{x} = x^{2}

x^{2} + \frac{1}{x^{2}} = 5

answer is D.

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Detailed Solution

To find the quadratic equation.
We know that, standard form of quadratic equation is,

a x^{2} + bx + c = 0

.
In option (1),

x^{3} - 3 \sqrt{x} + 2 = 0

Here, the equation is of 3 degrees, and as we know a quadratic equation is of 2 degrees.
Therefore, it is not a quadratic equation.
In option (2),

x + \frac{1}{x} = x^{2}

⟹ x^{2} + 1 = x^{3}

⟹ x^{2} = x^{3} - 1

⟹ x^{3} - x^{2} - 1 = 0

This equation has a degree of 3 and as we know, a quadratic equation is of 2 degrees.
Therefore, it is not a quadratic equation.
In option (3),

x^{2} + \frac{1}{x^{2}} = 5

⟹ x^{4} + 1 = 5 x^{2}

⟹ x^{4} - 5 x^{2} + 1 = 0

This equation has a degree of 4 and as we know, a quadratic equation is of 2 degrees.
Therefore, it is not a quadratic equation.
In option (4),

2 x^{2} - 5 x = {(x - 1)}^{2}

⟹ 2 x^{2} - 5 x = x^{2} - 2 x + 1

⟹ 2 x^{2} - 5 x - x^{2} + 2 x + 1 = 0

⟹ 2 x^{2} - x^{2} - 5 x + 2 x + 1 = 0

⟹ x^{2} - 3 x + 1 = 0

This equation has a degree of 2 and is in the standard form

a x^{2} + bx + c = 0

.
Therefore,

2 x^{2} - 5 x = {(x - 1)}^{2}

is a quadratic equation.
Hence, the correct option is 4.

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