Algebra Formulas

Algebra Formulas

Algebra is a branch of Mathematics that guides the students to solve mathematical equations and calculations of unknown quantities like percentages, constants, and variable values. The specific mathematical segment determines a situation in the presence of fixed and dynamic components.

Algebra is segregated into several categories: 1. Elementary Algebra, 2. Abstract Algebra, 3. Linear Algebra, 4. Advanced Algebra, and 5. Commutative Algebra. The secondary standard students need to understand Algebra formulas in-depth to solve complicated mathematical problems.

In the initial stages, the students find the Algebra formulas a little complicated and difficult to understand. Still, regular practice enables the students to deal with the algebraic problem with confidence readily. The students of all classes can download the algebraic formulas on Extramarks for free. 

Do You Know?

Algebra had its origin in Greek back in the 3rd Century. According to history, Babylonians created and developed algebra formulas and equations.

The modern concept of Algebra was brought to existence by Rene Descartes back in the 16th Century.

Sounds a bit exciting?

Yes! You will be surprised to know that we use the same formulas today in the 21st Century. Without the help of Algebra, digital mediums such as Google, the internet, smartphones, digital television, refrigerators, and other current technology might not have existed.

The study of Algebra revolves around terms, concepts, and formulas. The idea is very simple to the core. The mathematical symbols such as variables that imply the quantity leads to the solutions.

Algebraic properties:

• Addition’s Commutative property: a + b = b + a

If the order of the elements is modified, the sum of the expression does not change. Expressions or numbers can be used as elements.

• Multiplication’s Commutative Property: a x b = b x a

The product does not change when the order of the factors is changed. Numbers or phrases can be used as these factors.

• Addition’s Associative Property: (a + b)+ c = a + (b + c)

The property states that when two or more numbers are brought together to execute essential arithmetic addition, the order of the numbers has no bearing on the outcome.

• Multiplication has an associative property: (a x b) xc = a x (b x c)

When two or more factors are joined together in basic arithmetical multiplication, the order of the elements does not affect the final result. Also, in this situation, parenthesis is used to organise the items.

• Addition and Multiplication have distributive properties: 

a × (b + c) = a × b + a × c and (a + b) × c = a × c + b × c

The distributive property states that multiplying each element by a single term and then adding and subtracting the products is the same as multiplying each component by a single term and then adding and subtracting the products.

• Rule of multiplication over subtraction: p (q-r) = p*q – p*r 

If p, q, and r, are all integers. Likewise, you can use the left and right distributions in the addition rule for Multiplication over subtraction.

Left distributive law if p* (q-r) = (p * q) – (p*r)- and

Right distributive law if (p-q)*r = (p*r) – (q*r)-

Intricate knowledge of Algebra makes you think logically and solve complex mathematical problems efficiently. Algebraic identities are applicable in various branches of mathematics, including Algebra, Geometry, and Trigonometry.

Important Formulas in Algebra

The section has listed all algebra formulas to resolve the fundamental and complicated Mathematical problems for secondary standard students. 

The basic formulae  

1. a2 – b2 = (a – b)(a + b)
2. (a + b)2 = a2 + 2ab + b2
3. a2 + b2 = (a + b)2 – 2ab
4. (a – b)2 = a2 – 2ab + b2
5. (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
6. (a – b – c)2 = a2 + b2 + c2 – 2ab + 2bc – 2ca
7. (a + b)3 = a3 + 3a2b + 3ab2 + b3 ; (a + b)3 = a3 + b3 + 3ab(a + b)
8. (a – b)3 = a3 – 3a2b + 3ab2 – b3 = a3 – b3 – 3ab(a – b)
9. a3 – b3 = (a – b)(a2 + ab + b2)
10. a3 + b3 = (a + b)(a2 – ab + b2)
11. (a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4
12. (a – b)4 = a4 – 4a3b + 6a2b2 – 4ab3 + b4
13. a4 – b4 = (a – b)(a + b)(a2 + b2)
14. a5 – b5 = (a – b)(a4 + a3b + a2b2 + ab3 + b4)
15. (a +b+ c)2=a2+b2+c2+2ab+2bc+2ca
16. (a +b+ c+…)2=a2+b2+c2+⋯+2(ab +ac+ bc +⋯
17.  (x+ y+ z)2=x2+y2+z2+2xy+2yz+2xz
18. (x +y−z)2=x2+y2+z2+2xy−2yz−2xz
19. (x− y+ z)2=x2+y2+z2−2xy−2yz+2xz
20. (x−y−z)2=x2+y2+z2−2xy+2yz−2xz
21. x3+y3+z3−3xyz=(x+ y+ z)(x2+y2+z2−xy−yz−xz)
22. x2+y2=1/2[(x+ y)2+(x−y)2]
23. (x +a)(x +b)(x +c)=x3+(a +b+ c)x2+(ab +bc+ ca)x+ abc
24. x3+y3=(x+ y)(x2−xy+y2)
25. x3−y3=(x−y)(x2+xy+y2)
26. x2+y2+z2−xy−yz−zx=1/2[(x−y)2+(y−z)2+(z−x)2]

Laws of Exponents 

1. (am )(an )= am+n
2. am/an = am-n
3. (am)n = an
4. (ab) m = (am) ( bm)

Fractional Exponents

1. a0=1
2. am/an= am-n
3. am=1/a-m
4. a-m=1/am

If n is a natural number, the formula is:

1. an – bn = (a – b)(an-1 + an-2b+…+ bn-2a + bn-1)

If n is an even number, the formula is,

1.  (n = 2k), an + bn = (a + b)(an-1 – an-2b +…+ bn-2a – bn-1)

If n is an odd number, the formula would be,

1. (n = 2k + 1), an + bn = (a + b)(an-1 – an-2b +an-3b2…- bn-2a + bn-1)

Formulas of Quadratic Equation 

For a quadratic equation ax2 + bx + c = 0 where a ≠ 0, the roots will be given by the equation as x=−b±b2−4ac/2a

1. Δ = b2 − 4ac is called the discriminant
2. For real and distinct roots, Δ > 0
3. For real and coincident roots, Δ = 0
4. For non-real roots, Δ < 0
5. If α and β are the two roots of the equation ax2 + bx + c = 0 then, α + β = (-b / a) and α × β = (c / a).
6. If the roots of a quadratic equation are α and β, the equation will be (x − α)(x − β) = 0

Factorials 

1. n! = (1).(2).(3)…..(n − 1).n
2. n! = n(n − 1)! = n(n − 1)(n − 2)! = ….
3. 0! = 1
4. (a+b)n = an + nan-1b + n(n-1)/ 2! an-2b2 + n(n-1)(n-2)/2! An-3

Laws of Logarithm

1. xm = a⇒ logxa =m
2. loga1=0
3. logaa=1
4. loga(xy)=logax+logay
5. logax/y=logax– logay
6. loga(xm)=mlogax
7. loga =logcx/ logca
8. alog ax = x

Vector Algebra Formulas

Let P(x,y,z)P(x,y,z) be the point. The vector position of P is OP−→−=r⃗ =xı^+yȷ^+zk^OP→=r→=xı^+yȷ^+zk^ and, the magnitude of the vector is shown by |OP∣−→−=|r∣→=x2+y2+z2−−−−−−−−−−√|OP∣→=|r∣→=x2+y2+z2 

In case of any vector, rr is the magnitude, (l,m,n)(l,m,n) are the directions confines and (a,b,c)(a,b,c) are the directions ratios, then you see:

l=ar, m=br, n=er

Solved Examples

Sample 1

43 × 42 =?

Sol: Using the exponential formula (am) (an) = am+n where a = 4

= 43 × 42

= 43+2

= 45

= 1024

Sample 2

52 – 32 =?

Sol: Using the formula a2 – b2 = (a – b)(a + b)

where a = 5 and b = 3

= (a – b) (a + b)

= (5 – 3) (5 + 3)

= 2 × 8

= 16

Sample 3

Find the value of 42-22.

Sol: Using the formula a2–b2=(a–b)(a+b)a2–b2=(a–b)(a+b)

Where a=4a=4 and b=2b=2

= (a–b)(a+b)(a–b)(a+b)

= (4–2)(4+2)=(4–2)(4+2)

= 12

Sample 4

Find the value of (2x -3y) 2.

Sol: We use the identity here: (a–b) 2=a2–2ab+b2 to expand it 

Here, a= 2x and b=3y

We get (2x–3y)2 =(2x)2–2(2x)(3y)+(3y)2=4×2–12xy+9y2 

Hence, the required answer is (2x–3y)2=4×2–12xy+9y2

Sample 5

Solve, 4x + 5 when, x = 3.

Sol: Given, 4x + 5

Now putting the value of x=3, we get;

4 (3) + 5 

= 12 + 5

= 17.

Sample 6

12×2 – 9x + 5x – 4×2 – 7x + 10.

Sol: 12×2 – 9x + 5x – 4×2 – 7x + 10

= (12 – 4) x2 – 9x + 5x – 7x + 10

= 8×2 – 11x + 10

Sample 7

Solve (2x+y) 2

Sol: Using the identity: (a+b) 2 = a2 + b2 + 2 ab, we get;= (2x+y) = (2x) 2 + y2 + 2.2x.y = 4×2 + y2 + 4xy

Sample 8

({x^2} + 5x + 6 = 0\)

Sol: Here we are using algebra formulas for quadratic equations

We have x2+5x+6=0x2+5x+6=0

By comparing this with ax2+bx+c=0ax2+bx+c=0

We get a=1;b=5;c=6a=1;b=5;c=6 

Substitute these values in the quadratic formula:

x=–b±b2–4ac√2ax=–b±b2–4ac2a

x=–5±52–4⋅1⋅6√2⋅1x=–5±52–4⋅1⋅62⋅1

x=–5±1√2⋅1x=–5±12⋅1

x=–5±12,x=–5–12x=–5±12,x=–5–12

x=–2;x=–3x=–2;x=–3

Hence, x=–2x=–2 and −3

Sample 9

Let us now learn how to solve the algebra word problems by taking some examples.

A man starts his car from Delhi to Amritsar at 6.00 am. Assume the uniform speed of his car to be x km/h. At noon, he learns that he is 50 km away from Amritsar. Determine the distance between Delhi and Amritsar.

Sol: The time taken by the man to reach Amritsar = 12:00 noon – 6.00 am = 6 hours.

The uniform speed of the car = x km/h

Therefore, the total distance covered by the man = Time x speed = 6x km

Hence, the distance between Delhi and Amristar is (6x + 50) km. 

Sample 10

Find the value of y, when, 9y = 63

Sol: Divide both sides by 9;

y = 63/9

y = 7 

Sample 11

3×2 + 5 + 4×3 – x2 + 2×3 + 9 =?

Sol: =3×2 – x2 + 4×3 + 2×3 + 5 + 9 =

= 2×2 + 6×3 + 14 

Sample 12

5x + 15 = 65 

Sol: Learn how to isolate a variable.

=5x/5 + 15`/5 = 65/5

=x + 3 = 13 

= x 

= 10

Sample 13

My age: x

My brother is 3 years older than me: x + 3

My father is 3 less than 2 times my age: 2x – 3

My father’s age divided by 5 is equal to my brother’s age divided by 3: (2x – 3) / 5 = (x + 3) / 3

By cross multiplication:

5(x + 3) = 3(2x – 3)

5x + 15 = 6x – 9

x = 24

My father’s age: 2.24 – 3= 48 – 3 = 45

Sample 14

5(z + 1) = 3(z + 2) + 11. Z=?

Sol: 5z + 5 = 3z + 6 + 11

5z + 5 = 3z + 17

5z = 3Z + 17 – 5

5z – 3z = 12

2z = 12

Z = 12

Practice Questions

• If ‘a’ is the side-length of the equilateral triangle, then the perimeter of the triangle will be?
• Rahim is Radha’s elder brother, and Radha is 4 years younger than Rahim. Write about Leela’s age concerning Rahim’s age. Take Radha’s age to be x years.
• Give expressions for:

(i) p multiplied by 7

(ii) p divided by 7

• Milan’s age is x years old now. Then, what will be the age of Milan after 7 years?

1. 5x + 3 = 7x – 1. Find x
2. 5(z + 1) = 3(z + 2) + 11 Solve for Z
3. (x – 2) / 4 – (3x + 5) / 7 = – 3, x =?
4. 1 / (1 + 1 / (1 – 1/x)) = 4, x =?
5. 5x + 2(x + 7) = 14x – 7, x =?
6. 12t – 10 = 14t + 2, t=?

What to Expect from Extramarks?

Whether you are in class 6 or 9, or 12, algebra formulas are essential to have a clear idea of the subject and score high in Mathematics. Practicing Algebra is fun with Extramarks.

Students can download the algebra formula for free and study it at their convenience. The educational app makes the students understand the algebra formulas in-depth. It has adopted a unique way to encourage the students to solve mathematical problems with equations.

The solved questions and practice questions guide the students to practice regularly and analyze the subject better. The app is always there to support the students at every step when they face obstacles.

The academic experts have carefully made a list of all algebra formulas. All Extramarks does is make the students understand the complex problems readily. With Extramarks, the students of secondary standards enjoy the following benefits:

• Make the formulas simple to remember and implement.
• Perform and score better in all examinations
• Finish the Algebra syllabus before time
• Aware of the strengths and weaknesses
• Allocate extra time to improve the weaker sections
• Apart from Algebra, students become more comfortable with physics, engineering, chemistry, etc. 
• The students become confident with the performance
• The higher standard students can prepare for the high-level entrance examination.