(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}

**Solution:**

**(i) (x + 1)**^{2} = 2(x – 3)

^{2}= 2(x – 3)

By using the formula for (a+b)^{2 }= a^{2}+2ab+b^{2}

⇒ x^{2} + 2x + 1 = 2x – 6

⇒ x^{2} + 7 = 0

The above equation is in the form of ax^{2} + bx + c = 0

Therefore, the given equation is a quadratic equation.

### (ii) x^{2} – 2x = (–2) (3 – x)

⇒ x^{2 }– 2x = -6 + 2x

⇒ x^{2 }– 4x + 6 = 0

The above equation is in the form of ax^{2} + bx + c = 0

Therefore, the given equation is a quadratic equation.

### (iii) (x – 2)(x + 1) = (x – 1)(x + 3)

By multiplication,

⇒ x^{2 }– x – 2 = x^{2 }+ 2x – 3

⇒ 3x – 1 = 0

The above equation is not in the form of ax^{2} + bx + c = 0

Therefore, the given equation is not a quadratic equation.

### (iv) (x – 3)(2x +1) = x(x + 5)

By multiplication,

⇒ 2x^{2 }– 5x – 3 = x^{2 }+ 5x

⇒ x^{2 }– 10x – 3 = 0

The above equation is in the form of ax^{2} + bx + c = 0

Therefore, the given equation is a quadratic equation.

### (v) (2x – 1)(x – 3) = (x + 5)(x – 1)

By multiplication,

⇒ 2x^{2 }– 7x + 3 = x^{2 }+ 4x – 5

⇒ x^{2 }– 11x + 8 = 0

The above equation is in the form of ax^{2} + bx + c = 0.

Therefore, the given equation is a quadratic equation.

### (vi) x^{2} + 3x + 1 = (x – 2)^{2}

By using the formula for (a – b)^{2 }= a^{2 }– 2ab + b^{2}

⇒ x^{2} + 3x + 1 = x^{2} + 4 – 4x

⇒ 7x – 3 = 0

The above equation is not in the form of ax^{2} + bx + c = 0

Therefore, the given equation is not a quadratic equation.

### (vii) (x + 2)^{3} = 2x(x^{2} – 1)

By using the formula for (a + b)^{3 }= a^{3 }+ b^{3 }+ 3ab(a + b)

⇒ x^{3} + 8 + x^{2} + 12x = 2x^{3} – 2x

⇒ x^{3} + 14x – 6x^{2} – 8 = 0

The above equation is not in the form of ax^{2} + bx + c = 0

Therefore, the given equation is not a quadratic equation.

### (viii) x^{3} – 4x^{2} – x + 1 = (x – 2)^{3}

By using the formula for (a – b)^{3 }= a^{3 }– b^{3 }– 3ab(a – b)

⇒ x^{3} – 4x^{2} – x + 1 = x^{3} – 8 – 6x^{2 } + 12x

⇒ 2x^{2} – 13x + 9 = 0

The above equation is in the form of ax^{2} + bx + c = 0

Therefore, the given equation is a quadratic equation.