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# Check whether the following are Quadratic Equations. (i)$$(x + 1)^2 = 2(x - 3)$$ (ii)$$x^2 - 2x = (-2) (3 - x)$$ (iii)$$(x - 2) (x + 1) = (x - 1) (x + 3)$$ (iv)$$(x - 3) (2x + 1) = x (x + 5)$$ (v)$$(2x - 1) (x - 3) = (x + 5) (x - 1)$$ (vi)$$x^2 + 3x + 1 = (x - 2)^2$$ (vii)$$(x + 2)^3 = 2x(x^2 - 1)$$ (viii)$$x^3 - 4x^2 - x + 1 = (x - 2)^3$$

(i)$$(x + 1)^2 = 2(x - 3)$$
Using identity $$(a + b)^2 = a^2 + 2ab + b^2$$
$$\Rightarrow x^2 + 1 + 2x = 2x - 6$$
$$\Rightarrow x^2 + 7 = 0$$

Hence it is a quadratic equation since the degree is 2.

(ii)$$x^2 - 2x = (-2) (3 - x)$$
$$\Rightarrow x^2 - 2x = -6 + 2x$$
$$\Rightarrow x^2 - 2x - 2x + 6 = 0$$
$$\Rightarrow x^2 - 4x + 6 = 0$$

Hence it is a quadratic equation with degree 2.

(iii)$$(x - 2) (x + 1) = (x - 1) (x + 3)$$
$$\Rightarrow x^2 + x - 2x - 2 = x^2 + 3x - x - 3$$
$$\Rightarrow x^2 + x - 2x - 2 - (x^2 + 3x - x - 3) = 0$$
$$\Rightarrow x^2 - x^2 + x - 2x - 3x + x - 2 + 3 = 0$$
$$\Rightarrow -3x + 1 = 0$$

Hence it is a not quadratic equation since the degree is 1.

(iv)$$(x - 3) (2x + 1) = x (x + 5)$$
$$\Rightarrow 2x^2 + x - 6x - 3 = x^2 + 5x$$
$$\Rightarrow 2x^2 + x - 6x - 3 - x^2 - 5x = 0$$
$$\Rightarrow x^2 - 10x - 3 = 0$$

Hence it is a quadratic equation since the degree is 2.

(v)$$(2x - 1) (x - 3) = (x + 5) (x - 1)$$
$$\Rightarrow 2x^2 - x - 6x + 3 = x^2 - x + 5x - 5$$
$$\Rightarrow 2x^2 - 7x + 3 - x^2 + x - 5x + 5 = 0$$
$$\Rightarrow x^2 - 10x - 3 = 0$$

Hence it is a quadratic equation since the degree is 2.

(vi)$$x^2 + 3x + 1 = (x - 2)^2$$
Using identity $$(a - b)^2 = a^2 - 2ab + b^2$$
$$\Rightarrow x^2 + 3x + 1 = x^2 - 4x + 4$$
$$\Rightarrow x^2 + 3x + 1 - x^2 + 4x - 4 = 0$$
$$\Rightarrow 7x - 3 = 0$$

Hence it is a not quadratic equation since the degree is 1.

(vii)$$(x + 2)^3 = 2x(x^2 - 1)$$
Using identity $$(a + b)^3 = a^3 + b^3 + 3ab(a + b)$$
$$\Rightarrow x^3 + 2^3 + 3(x)(2)(x + 2) = 2x(x^2 - 1)$$
$$\Rightarrow x^3 + 8 + 6x(x + 2) = 2x^3 - 2x$$
$$\Rightarrow 2x^3 - 2x - x^3 - 8 - 6x^2 - 12x = 0$$
$$\Rightarrow x^3 - 6x^2 - 14x - 8 = 0$$

Hence it is a not quadratic equation since the degree is 3

(viii)$$x^3 - 4x^2 - x + 1 = (x - 2)^3$$
Using identity $$(a - b)^3 = a^3 - b^3 - 3ab(a - b)$$
$$\Rightarrow x^3 - 4x^2 - x + 1 = x^3 - 2^3 - 3(x)(2)(x-2)$$
$$\Rightarrow -4x^2 - x + 1 = -8 - 6x^2 + 12x$$
$$\Rightarrow 2x^2 - 13x + 9 = 0$$

Hence it is a quadratic equation since the degree is 2