# 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}

**Solution:**

Standard form of a quadratic equation is ax^{2} + bx + c = 0 in variable x where a, b, and c are real numbers and a ≠ 0. We need to check if the degree of the given equations is 2.

(i) (x + 1)^{2} = 2(x - 3)

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

x^{2} + 2x + 1 = (2x - 6)

x^{2} + 2x + 1 - (2x - 6) = 0

x^{2} + 2x + 1 - 2x + 6 = 0

x^{2} + 7 = 0

Here, the degree of x^{2} + 7 = 0 is 2.

∴ It is a quadratic equation.

(ii) x^{2} - 2x = (-2) (3 - x)

x^{2} - 2x = - 6 + 2x

x^{2} - 2x - 2x + 6 = 0

x^{2} - 4x + 6 = 0

Degree = 2

∴ It is a quadratic equation.

iii) (x - 2)( x + 1) = (x -1)( x + 3)

x^{2} - 2x + x - 2 = x^{2} + 3x - x - 3

x^{2} - x - 2 = x^{2} + 2x - 3

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

-3x + 1 = 0

Degree = 1

∴ It is not a quadratic equation.

iv) (x - 3)(2x +1) = x ( x + 5)

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

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

2x^{2} - 5x - 3 - x^{2} - 5x = 0

x^{2} - 10x - 3 = 0

Degree = 2

∴ It is a quadratic equation.

v) (2x -1)(x - 3) = (x + 5)(x -1)

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

2x^{2} - 7x + 3 = x^{2} + 4x - 5

2x^{2} - 7x + 3 - x^{2} - 4x + 5 = 0

x^{2} - 11x + 8 = 0

Degree = 2

∴ It is a quadratic equation

vi) x^{2} + 3x +1 = (x - 2)^{2}

x^{2} + 3x +1 = x^{2} - 4x + 4 [∵ (a - b)^{2} = a^{2} - 2ab + b^{2}]

x^{2} + 3x + 1 - x^{2} + 4x - 4 = 0

7x - 3 = 0

Degree = 1

∴ It is not a quadratic equation.

vii) (x + 2)^{3} = 2x (x^{2} -1)

We know that, (a + b)^{3} = a^{3} + 3a^{2}b + 3ab^{2} + b^{3}

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

x^{3} + 6x^{2} + 12x + 8 = 2x^{3} - 2x

- x^{3} + 6x^{2} + 14x + 8 = 0

Degree = 3

∴ It is not a quadratic equation.

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

x^{3} - 4x^{2} - x + 1 = x^{3} - 3(x)^{2}(2) + 3(x)(2)^{2} - (2)^{3} [∵ (a - b)^{3} = a^{3} - 3a^{2}b + 3ab^{2} - b^{3}]

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

x^{3} - 4x^{2} - x + 1 - x^{3} + 6x^{2} - 12x + 8 = 0

2x^{2} - 13x + 9 = 0

Degree = 2

∴ It is a quadratic equation

**☛ Check: **NCERT Solutions for Class 10 Maths Chapter 4

**Video Solution:**

## Check whether the following are quadratic equations: (i) (x + 1)² = 2(x - 3) (ii) x² - 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) x² + 3x +1 = (x - 2)²^{ }vii) (x + 2)³ = 2x (x² -1) viii) x³ - 4x² - x + 1 = (x - 2)³

NCERT Solutions Class 10 Maths Chapter 4 Exercise 4.1 Question 1

**Summary:**

The equations i) (x + 1)² = 2(x - 3) (ii) x² - 2x = (-2)(3 - x), ii) x² - 2x = (-2)(3 - x), iv) (x - 3)(2x +1) = x(x + 5), v) (2x -1)(x - 3) = (x + 5)(x -1) and viii) x³ - 4x² - x + 1 = (x - 2)³ are quadratic equations while iii) (x - 2)(x + 1) = (x - 1)(x + 3), vi) x² + 3x +1 = (x - 2)² and vii) (x + 2)³ = 2x (x² -1) are not quadratic equations.

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