# Quadratic Equation

## Quadratic Formula

`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, ax^{2} + bx + c = 0; the roots can be given by

#### Nature of Roots:

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

### NCERT Exercise 4.1

#### 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.