Introduction to Algebraic Formulas for Every Beginner:
Are you also frustrated with memorizing algebraic formulas like every student? If so, don’t worry; this article was written just to help students. 10 unique formulas are solved here, along with their derivations, relevant examples, and practical uses. You may easily learn square, cube, quadratic, and all other algebraic formulas. This article is especially useful for Class 8–10 students preparing for school and board exams. Let’s start the topic.
Square Formulas:
To solve each question, a trick is required that helps in solving it quickly. Square formulas are used in quadratic equations in the same way. By expanding these algebraic formulas, quadratic expressions are formed, and by factorizing quadratic expressions, squares are formed. We will derive the algebraic formulas given below one by one.
(a+b)2 = a2+2ab +b2
(a-b)2 = a2 -2ab +b2
a2-b2= (a+b) (a-b)
Derivation of Square Formulas
In numbers,
49 = 7^2 = (7) (7)
In variables, we can also write
(a+b)2= (a + b) (a + b)
(a+b)2 = a2 +2ab +b2
Step 1: Multiply
(a+b)2= (a + b) (a + b) = (a2 + b2 + ab + ba)
Step 2: Rearrange and add like terms
( a2 + ab + ba + b2 ) = ( a2+ ab + ab + b2 ) = ( a2 + 2ab + b2 )
So, this is how formulas are delivered.
An additional strategy for problem solving is to square the first and last integers, then multiply the product of the numbers by two and add them together. And if we move these around, nothing changes.
(a-b)2 = a2 -2ab +b2
Step 1: Multiply
(a-b)2= (a – b) (a – b) = ( a2– ab – ba + b2)
Step 2: Rearrange and add like terms
( a2– ab – ba + b2) = ( a2– ab – ab + b2) = ( a2– 2ab + b2)
If explained in simple words, then this formula is related to (a+b)2 = a2 +2ab +b2
Except here (-b) is used in place of b. So let’s show the same with proof.
2nd method:
Step 1: Multiply
(a+(-b))2= (a + (-b)) (a + (-b)) = ( a2– ab – ba + b2 )
Step 2: Rearrange and add like terms
( a2 – ab – ba + b2) = ( a2 – ab – ab + b2 ) = ( a2 – 2ab + b2 )
Extra Tip for Solving Problems:
An additional trick for problem-solving is to square the first and last integers, add them, multiply the product of the numbers by 2, and then subtract from the squares.
a2-b2= (a+b) (a-b)
Step 1: Multiply
(a+b) (a-b) = ( a2 – ab + ba – b2 )
Step 2: Cancel the opposite terms:
( a2 – ab + ba – b2 ) = a2– b2
Exam Tip:
Multiply the sum and difference of the numbers used in the question.
Algebraic Formulas Examples:
Here, 5 examples related to each type will be given so that it becomes easy for the students after practicing.
Identity 1
(a+b)2=a2+2ab+b2
Q1. Expand: (5x+1)2
(5x+1)2=(5x)2+2(5x)(1)+12
=25x2+10x+1
Q2 . Expand: (2y + 2)²
(2y + 2)² = (2y)² + 2(2y)(2) + 2²
= 4y² + 8y + 4
Q3. . Expand: (2a + 5b)²
(2a + 5b)² = (2a)² + 2(2a)(5b) + (5b)²
= 4a² + 20ab + 25b²
Q4. Find: (a + b)² if a = 3, b = 7
(a + b)² = a² + 2ab + b²
= 3² + 2(3)(7) + 7²
= 9 + 42 + 49
= 100
Q5. . Find: (x + y)² if x + y = 15
(x + y)² = (15)²
= 225
Identity 2
(a − b)² = a² − 2ab + b²
Q6 (Easy). Expand: (x − 3)²
(x − 3)² = (x)² − 2(x)(3) + 3²
= x² − 6x + 9
Q7. Expand: (5p − 1)²
(5p − 1)² = (5p)² − 2(5p)(1) + 1²
= 25p² − 10p + 1
Q8. Expand: (4a − 3b)²
(4a − 3b)² = (4a)² − 2(4a)(3b) + (3b)²
= 16a² − 24ab + 9b²
Q9. Find: (a − b)² if a = 10, b = 4
(a − b)² = a² − 2ab + b²
= 10² − 2(10)(4) + 4²
= 100 − 80 + 16
= 36
Q10. Find: (a − b)² if a² + b² = 41, ab = 20
(a − b)² = a² − 2ab + b²
=a² + b²− 2ab
= 41 − 40
= 1
Identity 3
a² − b² = (a + b)(a − b)
Q11. Factorize: x² − 9
x² − 9 = x² − 3²
= (x + 3)(x − 3)
Q12. Factorize: y² − 25
y² − 25 = y² − 5²
= (y + 5)(y − 5)
Q13. Factorize: 4a² − b²
4a² − b² = (2a)² − (b)²
= (2a + b)(2a − b)
Q14 . Find: a² − b² if a + b = 12, a − b = 4
a² − b² = (a + b)(a − b)
= 12 × 4
= 48
Q15. Simplify: (p² − q²) / (p − q)
(p² − q²) / (p − q)
= (p + q)(p − q) / (p − q)
= p + q
Derivation of Cube Formulas:
Cube means 3, it means that the formula having power of 3. Here, we will derive cubic formulas for your ease.
(a+b)3 = a3 +3a2b+3ab2 +b3
Step 1: Multiply
(a+b)3= (a + b)2 (a + b) = (a2 + 2ab+ b2)(a+b)
Step 2: Rearrange and add like terms
So, this is how formulas are delivered.
(a-b)3 = a3-3a2b+3ab2 -b3
Derivation of this algebraic formula is the same as that of the above algebraic formula. Just use -b in the place of b.
( a3 + a2b + 2a2b + 2ab2 +b2a+b3) = ( a3 + a2b + 2a2b + 2ab2 +ab2+b3) =a3+3a2b+3ab2 +b3
Step 1: Multiply
(a+(-b))3= (a + (-b))2 (a + (-b)) = (a2 – 2ab+ b2)(a-b)
Step 2: Rearrange and add like terms
( a3– a2b – 2a2b + 2ab2 + b2a – b3) = ( a3 – a2b – 2a2b + 2ab2 + ab2 – b3) = a3-3a2b+3ab2 -b3
Cube Formulas Examples
Question 1
Expand: (x+2)3
solution
(x+2)3=x3+3x2(2)+3x(22)+23
=x3+6x2+12x+8
Question 2
Expand: (a−3)3
Solution
(a−3)3=a3−3a2(3)+3a(32)−33
=a3−9a2+27a−27
Question 3
Expand: (2x+y)3
Solution
(2x+y)3=(2x)3+3(2x)2y+3(2x)y2+y3
=8x3+12x2y+6xy2+y3
Question 4
Expand: (m−n)3
Solution
=m3−3m2n+3mn2−n3
=m3−3m2n+3mn2−n3
Question 5
Expand: (x−1)3
Solution
=x3−3x2(1)+3x(1)−1
= x3−3x2+3x−1
Question 6
Expand: (3a+b)3
Solution
=(3a)3+3(3a)2b+3(3a)b2+b3
=27a3+27a2b+9ab2+b3
Question 7
Expand: (2x−5)3
Solution
=(2x)3−3(2x)2(5)+3(2x)(52)−53
= 8x3−60x2+150x−125
Question 8
Expand: (a+4b)3
Solution
=a3+3a2(4b)+3a(4b)2+(4b)3
Question 9
Expand: (y−2x)3
Solution
=y3−3y2(2x)+3y(2x)2−(2x)3
Question 10
Expand: (p+q)3
Solution
=p3+3p2q+3pq2+q3
Quadratic Formulas
Using this formula, the quadratic equation is factorized. We will determine this formula and work through some associated algebraic problems in this part.
Let’s have a quadratic equation
ax2 + bx + c = 0
Divide this equation by a:
x²+b/a x+ca= 0
x²+b/a x=-ca
For Completing Square, add b24a on both sides,
x²+b/a x+ b²/4a=-ca+b²/4a
(x+b/2a )²=b²-4ac/4a²
Taking the square root on both sides,
√(x+b/2a )²=√b²-4ac/4a²
x=-b/2a+√ b2-4ac/4a2
By taking LCM:
x=-b±√b2-4ac / 2a This is the quadratic formula.
Solved Algebraic Problems:
1. Solve x2−5x+6=0
Solution:
Here a = 1, b = -5 and c = 6
By using the Quadratic Formula:
x= -b±√b2-4ac / 2a
x= -(-5)±√(-5)2-4(1)(6) / 2(1)
x= 5±√25-24 / 2
x= 5±√1 / 2
x= 5+1 / 2 or x= 5 – 1 / 2
x= 62 or x= 42
x = 3 or x= 2
2. Solve 2x2+3x−7=0
Solution:
Here a = 2, b = 3 and c = -7
By using the Quadratic Formula:
x= -b±√b2 -4ac /2a
x= -3±√(3)2-4(2)(-7)/2(2)
x= -3±√9+56 / 4
x= -3±√65 / 2
x = -3+√65 / 2 or -3-√65 / 2
3. Solve x2+4x+9=0
Solution:
Here a = 1, b = 4 and c = 9
By using the Quadratic Formula:
x= -b±√b2 -4ac /2a
x= -4±√(4)2-4(1)(9)/2(1)
x= -4±√16-36/4
x= -3±√-20/2
Algebraic Problems for Practice:
Some problems related to all the above formulas are given. You should solve these.
- Expand:
(2x+3)2
- Expand:
(3a−5b)2
3. Expand:
(x+2y)3
4 Factorize:
x2−16
5. Solve by using the quadratic formula:
2x2+5x−3=0
Formulas on Wikipedia are a good place to look up further information regarding algebraic formulas.
Other Algebraic Formulas List:
Here are some algebraic formulas that are useful for every student at a medium level.
- (a+b+c)2=a2+b2+c2+2(ab+bc+ca)
- (a−b−c)2=a2+b2+c2−2ab−2ac+2bc
x2+(a+b)x+ab=(x+a)(x+b)- x2−(a+b)x+ab=(x−a)(x−b)
- a0=1
- a−n=1/an
- am⋅an=am+n
- am/an=am-n
Difference Between Algebraic Expressions and Identities:
| Features | Algebraic Expressions | Identites |
| Does it contain an equality sign? | No | Yes |
| Always true? | Not necessarily | Yes |
For a deeper understanding, you can also review the algebra articles for classes 6, 7, and 8.
FAQ’s
Q1. Describe a simple algebraic formula.
E=mc2 is the fundamental algebraic formula.
Q2 Are algebraic formulas difficult to learn?
No, it won’t be hard if you fully comprehend it.
Q3. What errors do students frequently make?
When it comes to posting signage, students typically make mistakes.
Conclusion:
The fundamental algebraic formulas and their derivations were taught to us in this article. We have additionally solved problems, and from the viewpoint of the test, these formulas are essential for all students. Any student who learns these formulas will receive maximum marks on the assessment.
