ax^2 + bx + c = 0 is a quadratic equation in the variable x. Here a, b, c are real numbers and a ≠ 0

If α is the root of quadratic equation ax^2 + bx + c = 0

Then, aα^2+bα+c=0

For a given quadratic equation, ax2 + bx + c = 0; the roots can be given by

(-b±sqrt(b^2-4ac))/(2a)

#### Nature of Roots:

1. If b^2 – 4ac > 0, the equation has two distinct roots.
2. If b^2 – 4ac = 0, the equation has two equal roots.
3. If b^2 – 4ac < 0, the equation has no real roots.

### Exercise 4.1 (NCERT Solution) - Part - 1

Check whether the following are quadratic equations:

(i) (x + 1)^2 = 2(x – 3)

Solution: LHS: (x + 1)^2 = x^2 + 2x + 1

(Using (a + b)^2 = a^2 + 2ab + b^2)

RHS: 2(x – 3) = 2x – 6

Now; x^2 + 2x + 1 = 2x – 6

Or, x^2 + 2x + 1 – 2x + 6 = 0

Or, x^2 + 7 = 0

Since the equation is in the form of ax^2 + bx + c = 0; hence it is a quadratic equation.

(ii) x^2 – 2x = (-2)(3 – x)

Solution: Solution: x^2 – 2x = (-2)(3 – x)

Or, x^2 – 2x = -6 – 2x

Or, x^2 – 2x + 2x + 6 = 0

Or, x^2 + 6 = 0

Since the equation is in the form of ax^2 + bx + c = 0; hence it is a quadratic equation.

(iii) (x – 2)(x + 1) = (x – 1)(x + 3)

Solution: Solution: LHS: (x – 2)(x + 1)

= x^2 + x – 2x – 2

= x^2 – x – 2

RHS: (x – 1)(x + 3)

= x^2 + 3x – x + 3

= x^2 + 2x + 3

Now; x^2 – x – 2 = x^2 + 2x + 3

Or, x^2 – x – 2 – x^2 – 2x – 3 = 0

Or, - 3x – 5 = 0

Since the equation is not in the form of ax^2 + bx + c = 0; hence it is not a quadratic equation.

(iv) (x – 3)(2x + 1) = x(x + 5)

Solution:LHS: (x – 3)(2x + 1)

= 2x^2 + x – 6x – 6

= 2x^2 – 5x – 6

RHS: x(x + 5)

= x^2 + 5x

Now; 2x^2 – 5x – 6 = x^2 + 5x

Or, 2x^2 – 5x – 6 – x^2 – 5x = 0

Or, x^2 – 10x – 6 = 0

Since the equation is in the form of ax^2 + bx + c = 0; hence it is a quadratic equation.

(v) (2x – 1)(x – 3) = (x + 5)(x – 1)

Solution: LHS: (2x – 1)(x – 3)

= 2x^2 – 6x – x + 3

= 2x^2 – 7x + 3

RHS: (x + 5)(x – 1)

= x^2 – x + 5x – 5

= x^2 + 4x – 5

Now; 2x^2 – 7x + 3 = x^2 + 4x – 5

Or, 2x^2 – 7x + 3 – x^2 - 4x + 5 = 0

Or, x^2 – 11x + 8 = 0

Since the equation is in the form of ax^2 + bx + c = 0; hence it is a quadratic equation.

(vi) x^2 + 3x + 1 = (x – 2)^2

Solution: x^2 + 3x + 1 = (x – 2)^2

Or, x^2 + 3x + 1 = x^2 – 4x + 4

Or, x^2 + 3x + 1 – x^2 + 4x – 4 = 0

Or, 7x – 3 = 0

Since the equation is not in the form of ax^2 + bx + c = 0; hence it is not a quadratic equation.

(vii) (x + 2)^3 = 2x(x^2 – 1)

Solution: LHS: (x + 2)^3

Using (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3, we get;

(x + 2)^3 = x^3 + 6x^2 + 12x + 8

RHS: 2x(x^2 – 1)

= 2x^3 – 2x

Now; x^3 + 6x^2 + 12x + 8 = 2x^3 – 2x

Or, x^3 + 6x^2 + 12x + 8 – 2x^3 + 2x = 0

Or, - x^3 + 6x^2 + 14x + 8 = 0

Since the equation is not in the form of ax^2 + bx + c = 0; hence it is not a quadratic equation.

(viii) x^3 – 4x^2 – x + 1 = (x – 2)^3

Solution: Solution: LHS: x^3 – 4x^2 – x + 1

RHS: (x – 2)^3

= x^3 – 8 – 6x^2 + 12x

Now; x^3 – 4x^2 – x + 1 = x^3 – 6x^2 + 12x – 8

Or, x^3 – 4x^2 – x + 1 – x^3 + 6x^2 – 12x + 8 = 0

Or, 2x^2 – 13x + 9 = 0

Since the equation is in the form of ax^2 + bx + c = 0; hence it is a quadratic equation.