Introduction to identities — lesson. Mathematics State Board, Class 7.

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In previous classes, we have learnt to derive algebraic expressions using exponential notations.

Consider an algebraic expression

x^{2} + 5x + 4 = 0

with the variable \(x\). This can also be written as an equation

x^{2} + 5x = - 4

We can verify the equation by substituting numerical values of \(x\).

If \(x = -4\) then:

\begin{array}{l} LHS = x^{2} + 5x = ({- 4}^{2}) + 5 (- 4) \\ = 16 - 20 \\ = - 4 = RHS \end{array}

This equation is true when \(x = -4\). But this equation is not true when \(x = -2\).

Let's substitute \(x = -2\) to find out the result.

\begin{array}{l} LHS = x^{2} + 5x = ({- 2}^{2}) + 5 (- 2) \\ = 4 - 10 \\ LHS = - 6 \neq RHS \end{array}

From the above two examples, we can observe the equation

x^{2} + 5x + 4 = 0

is true when \(x\) takes certain values such as \(-4\) but not \(-2\).

Therefore, an equation is true only for certain values of the variable in it. It is not true for all values of the variables.

So, is there any algebraic equation that holds true for all values of the variables? Yes, there is an equation that is called identities.

Identities:

Let us the algebraic expression,

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

. Let us try to find the values of the expression for the given values of \(a\) and \(b\).

Let us take \(a =\)6 and \(b =\)10.

{LHS = (a + b)}^{2} = {(6 + 10)}^{2} = {(16)}^{2} = 256

\begin{array}{l} RHS = a^{2} + 2 ab + b^{2} = 6^{2} + 2 (6 \times 10) + 10^{2} \\ = 36 + 2 (60) + 100 \\ = 36 + 120 + 100 \\ RHS = 256 \end{array}

We can see that \(LHS = RHS =\) 256. This means if we substitute any numeric variables as \(a\) and \(b\), this equation will be true, \(LHS\) and \(RHS\) will be the same. When an equation gives such equality, it is called identity.

Therefore, the equation

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

is a identity.

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1. Introduction to identities

Theory: