Algebraic Identities

An algebraic formula is an equation that is a rule written using mathematical and algebraic symbols. 

It is an equation that involves algebraic expressions on both sides.

Example:

\((a+b)^{2}=a^{2}+2 a b+b^{2}\) is an algebraic formula and here,

• \((a+b)^{2}\) is an algebraic expression.
• \(a^{2}+2 a b+b^{2}\) is an algebraic expression.

The algebraic formulas that are involved in class 8 are:

Algebraic Formulas \[ \begin{align} (a+b)^2 &= a^2+2ab+b^2\\[0.3cm] (a-b)^2 &=a^2-2ab+b^2 \\[0.3cm] a^2-b^2&= (a+b)(a-b)\\[0.3cm] (a+b)^3 &= a^3+3a^2b+3ab^2+b^3\\[0.3cm] (a-b)^3 &= a^3-3a^2b+3ab^2-b^3\\[0.3cm] a^3+b^3&= (a+b) (a^2-ab+b^2)\\[0.3cm] a^3-b^3&= (a-b) (a^2+ab+b^2)\\[0.3cm] (a+b+c)^2 &= a^2+b^2+c^2+2ab+2bc+2ca \end{align} \]

Some laws of exponents are:

Laws of Exponents \[ \begin{align} a^m \cdot a^n& = a^{m+n}\\[0.3cm] \frac{a^m}{a^n} &= a^{m-n}\\[0.3cm] (a^m)^n &= a^{mn}\\[0.3cm] (ab)^m &= a^m \cdot b^m \\[0.3cm] a^0 &=1\\[0.3cm] a^{-m} &= \frac{1}{a^m} \end{align} \]

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"Logarithms" is introduced in class 9 and here are some formulas related to this concept.

Properties of Logarithms \[\begin{align} x^m =a &\Rightarrow \log_x a = m\\ \log _{a} 1 &=0\\ \log _{a} a &=1\\ \log _{a}(xy) &=\log _{a} x+\log _{a} y\\ \log _{a} \frac{x}{y} &=\log _{a} x-\log _{a} y\\ \log _{a}\left(x^{m}\right) &=m \log _{a} x\\ \log _{a} x &=\frac{\log _{c} x}{\log _{c} a}\\ a^{\log _{a} x} &=x \end{align}  \]

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Algebra Formulas for Class 10

An important algebra formula in class 10 is the “quadratic formula”. 

The “quadratic formula” is used to find the roots of a quadratic equation \(ax^2+bx+c=0\).

This formula says:

\[ x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a} \]

Here:

• If \(b^2-4ac>0\) then the quadratic equation has two distinct real roots.
• If \(b^2-4ac=0\) then the quadratic equation has two equal real roots.
• If \(b^2-4ac<0\) then the quadratic equation has two imaginary roots.

Apart from this, we have a few other formulas related to progressions.

1) nth term of an arithmetic sequence;	\({a_n} = a+(n-1)d\)
2) Sum of n terms of an arithmetic sequence;	\(S_n = \frac{n}{2} (2a+(n-1)d)\)
3) nth term of a geometric sequence;	\(a_n = a \cdot r^{n-1}\)
4) Sum of n terms of a geometric sequence;	\(S_n = \frac{a\left(1-r^{n}\right)}{1-r}, r \neq 1\)
5) Sum of infinite terms of a geometric sequence;	\(S = \frac{a}{1-r}\)

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The algebraic formulas that are involved in class 11 are as follows:

Algebraic Formulas \[ \begin{align} n! &= n \times (n-1) \times …. 3 \times 2 \times 1\\[0.3cm] _{n} C_{r}&=\frac{n !}{(n-r) ! r !}\\[0.3cm] _{n} P_{r}&=\frac{n !}{(n-r) !} \end{align} \]

Apart from these, there is “Binomial Theorem” as well which is used to evaluate the large exponents of algebraic expressions with two terms. It says:

The vector algebra formulas that are involved in class 12 are as follows.

For any three vectors, \( \vec{a}, \vec{b}\) and \(\vec{c}\):

• The magnitude of \(\vec{a} = x \hat{i}+y \hat{j}+z \hat{k} \) is, \( |\vec{a}| = \sqrt{x^2+y^2+z^2} \).
   
• The unit vector along \(\vec{a}\) is \(\dfrac{\vec{a}}{|\vec{a}|}\).
   
• The dot product is defined as: \(\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta\), where \(\theta\) is the angle between the vectors \(\vec{a}\) and \(\vec{b}\).
   
• The cross product is defined as: \(\vec{a} \times \vec{b} = |\vec{a}| |\vec{b}| \sin \theta \,\,\hat{n}\), where \(\theta\) is the angle between the vectors \(\vec{a}\) and \(\vec{b}\).
   
• The scalar triple product of three vectors is given by \( [\vec{a} \text{  } \vec{b} \text{  } \vec{c} ] = \vec{a} \cdot (\vec{b} \times \vec{c}) = (\vec{a} \times \vec{b}) \cdot \vec{c} \).

An algebraic function is of the form \(y=f(x)\), where 

• \(x\) is called the input.
• \(y\) is called the output.

Here, each input corresponds to exactly one output.

But multiple inputs may correspond to a single output.

For example:

\(f(x)= x^2\) is an algebraic function.

Here, when \(x=2\), \(f(2)= 2^2=4\).

Here, \(x=2\) is the input and \(f(2)=4\) is the output of the function.

We can add, subtract, multiply, and divide fractions in algebra just the same way we do with fractions involving numbers.

Adding Fractions

\[ \frac{x}{y}+\frac{z}{w}=\frac{x w+y z}{y w} \]

Subtracting Fractions

\[ \frac{x}{y}-\frac{z}{w}=\frac{x w-y z}{y w} \]

Multiplying Fractions

\[ \frac{x}{y} \times \frac{z}{w}=\frac{x z}{y w} \]

Dividing Fractions

\[ \frac{x}{y} \div \frac{z}{w}=\frac{x w}{y z} \]

Example 1

 

 

Using algebraic identities, find \((2x-3y)^2\).

Solution:

Here, we use the identity \((a-b)^2= a^2-2ab+b^2\) to expand this.

Here, \( a= 2x\) and \(b=3y\).

Then we get:

\[ \begin{align}
(2x-3y)^2 &= (2x)^2 -2(2x)(3y)+(3y)^2 \\
&= 4x^2 -12xy+9y^2
\end{align} \]

\(\therefore (2x-3y)^2=4x^2 -12xy+9y^2\)

Example 2

 

 

Using identities, evaluate \(297 \times 303\). 

Solution:

The above product can be written as \((300-3) \times (300+3)\).

We will find this product using the formula:  

\[(a-b)(a+b)=a^{2}-b^{2} \]

Here \(a=300\) and \(b=3\).

Then we get:

\[ \begin{aligned}
(300-3) \times(300+3) &=300^2-3^2 \\
&=90000-9 \\
&=89991
\end{aligned} \]

\(\therefore 297 \times 303= 89991\)

Example 3

 

 

Find the roots of the quadratic equation \(x^2+5x+6=0\) using the quadratic formula.

Solution:

The given equation is \(x^2+5x+6=0\).

Comparing this with \(ax^2+bx+c=0\), we get:

\[a=1; \,\, b=2; \,\, c=2\].

Substituting these values in the quadratic formula:

\[ \begin{align}
x&=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\\[0.2cm]
x&=\frac{-5 \pm \sqrt{5^2-4 \cdot 1 \cdot 6}}{2 \cdot 1}\\[0.2cm]
x&=\frac{-5 \pm \sqrt{1}}{2 \cdot 1}\\[0.2cm]
x&=\frac{-5 \pm 1}{2}\\[0.2cm]
x &= \frac{-5+1}{2}, \,\,\, x=\frac{-5-1}{2}\\[0.2cm]
x &= -2; \,\,\, x=-3
\end{align} \]

\(\therefore x= -2, -3\)

Example 4

 

 

Use the laws of exponents and simply the following expression:

\[ \left(\frac{6 \times 10}{2^{2} \times 5^{3}}\right)^{2} \times \frac{25}{27} \]

Solution:

We will simplify the given expression using the laws of exponents:

\begin{aligned} \left(\frac{6 \times 10}{2^{2} \times 5^{3}}\right)^{2} \times \frac{25}{27} &=\left(\frac{2 \times 3 \times 2 \times 5}{2^{2} \times 5^{3}}\right)^{2} \times \frac{5^{2}}{3^{3}} \quad[\because 6=2 \times 3 \text { and } 10=2 \times 5] \\[0.3cm] &=\left(\frac{2^{2} \times 3 \times 5}{2^{2} \times 5^{3}}\right)^{2} \times \frac{5^{2}}{3^{3}} \\[0.3cm]\ &=\left(\frac{3}{5^{2}}\right)^{2} \times \frac{5^{2}}{3^{3}} \left [\because(a \times b)^{m}=a^{m} \times b^{m}\right] \\[0.3cm] &=\frac{3^{2}}{5^{4}} \times \frac{5^{2}}{3^{3}} \\[0.3cm] &=\frac{1}{5^{2} \times 3} \\[0.3cm] &=\frac{1}{25 \times 3} \\[0.3cm] &=\frac{1}{75} \quad\left[\because\left(a^{m}\right)^{n}=a^{m n}\right] \end{aligned}

\( \therefore  \left(\frac{6 \times 10}{2^{2} \times 5^{3}}\right)^{2} \times \frac{25}{27} = \dfrac{1}{75} \)

Example 5

 

 

Compress the following expression into a single logarithm using the properties of logarithms:

\[ 5 \log _{3}(x)+2 \log _{3}(A x)-\log _{3}\left(8 x^{5}\right)\]

Solution:

We will compress the given expression using the properties of logarithms:

\begin{align}5 \log _{3}(x)+2 \log _{3}(4 x)-\log _{3}\left(8 x^{5}\right) &=\log _{3}\left(x^{5}\right)+\log _{3}\left( (4 x)^{2}\right)-\log _{3}\left(8 x^{5}\right) \\[0.3cm] &=\log _{3}\left(x^{5}\right)+\log _{3}\left(16 x^{2}\right)-\log _{3}\left(8 x^{5}\right) \\[0.3cm] &=\log _{3}\left(x^{5} \cdot 16 x^{2}\right)-\log _{3}\left(8 x^{5}\right) \\[0.3cm] &=\log _{3}\left(16 x^{7}\right)-\log _{3}\left(8 x^{5}\right) \\[0.3cm] &=\log _{3}\left(\frac{16 x^{7}}{8 x^{5}}\right)\\[0.3cm] &= \log _{3}\left(2 x^{2}\right) \end{align}

\( \therefore 5 \log _{3}(x)+2 \log _{3}(A x)-\log _{3}\left(8 x^{5}\right) = \log _{3}\left(2 x^{2}\right) \)

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