Which of the following equations are quadratic equations?

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

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

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(5 x^{2} + 2 x) - (7 + 5 x^{2}) = x^{2} + 2

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(x + 2)^{3} = 4 - x^{3}

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Solution

The correct options are
A $(x^{2} + 1)^{2} = x^{4}$
C $(5 x^{2} + 2 x) - (7 + 5 x^{2}) = x^{2} + 2$
The standard form of a quadratic equation is given by $a x^{2} + b x + c = 0,$ provided $a, b$ and $c$ are real numbers where $a \neq 0.$
So, let us simplify the given equations to check if they are quadratic or not.

Consider $(x^{2} + 1)^{2} = x^{4} .$
$\Rightarrow x^{4} + 2 x^{2} + 1 = x^{4}$
$\Rightarrow 2 x^{2} + 1 = 0,$ a quadratic equation.

Simplifying $(x + 1)^{2} = x^{2}$
$\Rightarrow x^{2} + 2 x + 1 = x^{2}$
$\Rightarrow 2 x + 1 = 0,$ a linear equation (and hence not a quadratic equation).

Simplifying $(5 x^{2} + 2 x) - (7 + 5 x^{2}) = x^{2} + 2$
$\Rightarrow 2 x - 7 = x^{2} + 2$
$\Rightarrow x^{2} - 2 x + 9 = 0,$ a quadratic equation.

Lastly, let us consider $(x + 2)^{3} = 4 - x^{3} .$
$\Rightarrow x^{3} + 6 x^{2} + 12 x + 8 = 4 - x^{3}$
$\Rightarrow 2 x^{3} + 6 x^{2} + 12 x + 4 = 0,$ a cubic equation as the highest degree of $x$ in the given equation is 3.

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Similar questions

Check whether the following are quadratic equations :
(i) ${(x + 1)}^{2} = 2 (x - 3)$
(ii) $x^{2} - 2 x = (- 2) (3 - x)$
(iii) $(x - 2) (x + 1) = (x - 1) (x + 3)$
(iv) $(x - 3) (2 x + 1) = x (x + 5)$
(v) $(2 x - 1) (x - 3) = (x + 5) (x - 1)$
(vi) $x^{2} + 3 x + 1 = {(x - 2)}^{2}$
(vii) ${(x + 2)}^{3} = 2 x (x^{2} - 1)$
(viii) $x^{3} - 4 x^{2} - x + 1 = {(x - 2)}^{3}$