# Equation and Identity:

#### Algebraic Equation:

If an algebraic expression is equal to zero then it is called an algenraic equation.

Example: $$f(x) = 4x^2 + 2x + 1$$, the given expression is an algebraic function and if we put this equation equal to zero then it is called an algebraic equation. $$4x^2 + 2x + 1 = 0$$ this is an algebraic equation.

Note: The certain values of variable $$x$$, which can satisfy the equation are called the roots of that equation.

#### Algebraic Identity:

The algebraic equation and identity are same, the only difference is an equation can be satisfied by some certain values of variable $$x$$ whereas an identity can be satisfied by all values of variable $$x$$.

Hence an equation which satisfy by all the values of variable $$x$$ is called an identity.

#### Factor Theorem:

According to the factor theorem if "$$\alpha$$" is a root of an equation $$f(x) = 0$$ then the polynomial $$f(x)$$ will be completely divisible by $$(x - \alpha)$$ or we can say $$(x - \alpha)$$ will be a factor of polynomial equation $$f(x)$$.

#### Proof:

If $$\alpha$$ is a root of $$f(x) = 0$$ then it will satify the equation $$f(x) = 0$$. $$f(\alpha) = 0.....(1)$$ Let by dividing $$f(x)$$ from $$(x - \alpha)$$ we get remainder $$R$$ and quotient $$k(x)$$ then

dividend = divisor $$\times$$ quotient + remainder $$f(x) = (x - \alpha) \times k(x) + R.....(2)$$ By putting $$x = \alpha$$ in equation (2) we get $$f(\alpha) = (\alpha - \alpha) \times k(\alpha) + R$$ $$f(\alpha) = R$$ from equation (1) $$f(\alpha) = 0$$ So $$f(\alpha) = R = 0$$ Here, it is clear that by dividing $$f(x)$$ from $$(x - \alpha)$$ we get remainder zero hence $$x - \alpha$$ is a facror of polynomial equation $$f(x)$$.

#### Roots of two degree equation:

The roots of two degree equation can be find out by the given formula. $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ Here $$(b^2 - 4ac)$$ is called discriminant because the roots of the equation are dependent of discriminant.

Notes: (1): For an $$n$$ degree equation, the number of roots will also be $$n$$.

(2): For algebraic equation with real factors, if one root is $$(\alpha + i \beta)$$ then second root must be $$(\alpha - i \beta)$$.

(3): For algebraic equation with rational factors, if one root is $$(\alpha + \sqrt{\beta})$$ then the second root must be $$(\alpha - \sqrt{\beta})$$.