Algebraic Identities For Class 8

Algebraic Identities

We define identity as an equality which is true for all values of the variable. These identities are the algebraic identities, which clearly define that the (LHS) and (RHS) of the equation is equal for all the values of the variable. Algebraic expressions are usually expressed as monomials, binomials and trinomials. This description is based on the fact that how many terms are present in the expression. It may be one, two, or three. In fact, the expression which has one or more than one terms present in it is called a polynomial and the number attached to the term is called a coefficient.

The algebraic identities class 8 consist of three major identities.

These are very basic and general algebraic identities. Substituting the values for a and b, in any of them, the left-hand side of the equation will be equal to the right-hand side. Therefore, these expressions are called as identities.

All Algebraic Identities Class 8

Identity 1: $(a + b) ^ 2 = a ^ 2 + 2ab + b ^ 2$

Proof: Lets start with left hand side,

$(a + b) ^ 2 = (a + b)(a + b)$

By distributive law;

$(a + b) ^ 2 = a(a + b) + b(a + b)$

By multiplying each term, we get,

$(a + b) ^ 2 = a ^ 2 + ab + ba + b ^ 2$

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

$L.H.S. = R.H.S.$

Identity 2: $(a - b) ^ 2 = a ^ 2 - 2ab + b ^ 2$

Proof: Lets start with left hand side,

$(a - b) ^ 2 = (a - b)(a - b)$

By distributive law;

$(a - b)² = a (a - b) - b (a - b)$

By multiplying each term, we get,

$(a - b) ^ 2 = a ^ 2 - ab - ba + b ^ 2$

$(a - b) ^ 2 = a ^ 2 - 2ab + b ^ 2$

$L.H.S. = R.H.S.$

Identity 3: $(a + b)(a - b) = a ^ 2 - b ^ 2$

Proof: Starting with left hand side, by distributive law;

$(a + b) (a - b) = a(a - b)+b(a - b)$

Multiplying each term, we get,

$(a + b)(a - b) = a ^ 2 - ab + ab - b ^ 2$

$(a + b)(a - b) = a ^ 2 - b ^ 2$

$L.H.S. = R.H.S.$

Hence, we have successfully proved all algebraic identities class 8.

Algebraic Identities Class 8 Extra Questions

Now let us look into some algebraic identities class 8 questions and answers.

Question 1: Solve $(4 x+2)(4 x-2)$ using algebraic identities.

Solution: We can write the given expression as:

$(4 x+2)(4 x-2)=(4 x)^2-(2)^2=16 x^2-4$

Question 2: Solve $(3 x+6)^2$ using algebraic identities.

Solution: We write the given expression as:

$\begin{aligned} & (3 x+6)^2=(3 x)^2+2^* 3 x^* 6+6^2 \\ & (3 x+5)^2=9 x^2+12 x+36\end{aligned}$

Question 3: Expand $(2 \mathrm{x}+2 \mathrm{y})^2$.

Solution: To expand the given expression, we substitute $a=2 x$ and $b=2 y$ in $(a+$ b) ${ }^2=a^2+2 a b+b^2$,

$
\begin{aligned}
& (2 x+2 y)^2=(2 x)^2+2(2 x)(2 y)+(2 y)^2 \\
& =4 x^2+8 x y+4 y^2
\end{aligned}
$

Question 4: Using algebraic identities for class 8, solve $296 \times 304$.

Solution: $296 × 304$ can be written as $( 300 - 4 ) \times ( 300 + 4 )$

And this is based on the algebraic identity $(a+b)(a-b)=a^2-b^2$

Here we have $a=300$, and $b=4$

Substituting the values in the above identity, we get:

$
\begin{aligned}
& (300-4)(300+4)=300^2-4^2 \\
& =90000-16 \\
& =89984
\end{aligned}
$

Question 5: Simplify $( 7x + 2y )^2 + ( 7x - 2y )^2$

Solution: To solve this, we need to use the following algebra identities:

$
\begin{aligned}
& (a+b)^2=a^2+2 a b+b^2 \\
& (a-b)^2=a^2-2 a b+b^2
\end{aligned}
$

Adding the above two formulas we have:

$
\begin{aligned}
& (a+b)^2+(a-b)^2=a^2+2 a b+b^2+a^2-2 a b+b^2 \\
& (a+b)^2+(a-b)^2=2 a^2+2 b^2
\end{aligned}
$

Here we have $\mathrm{a}=7 \mathrm{x}$ and $\mathrm{b}=2 \mathrm{y}$. Substituting this in the above expression we have:

$
\begin{aligned}
& (7 x+2 y)^2+(7 x-2 y)^2=2(7 x)^2+2(2 y)^2 \\
& =98 x^2+8 y^2
\end{aligned}
$

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