Algebraic formulas and identities are useful for solving equations quickly. But it is difficult to memorize the formulas without knowing their derivation. Here are the important algebraic formulas and identities derivations with examples. After this article, every beginner can derive a formula and easily use it in solving questions. This derivation of algebraic formulas and identities will be helpful for students in classes 6 to 8.
Related Topics:
Algebraic Equations with answers
Algebraic Expressions for Beginners
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.
Important Algebraic Formulas and Identities 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 has a 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 b2/4a 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 important algebraic formulas and identities 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 |
FAQ’s
Q1. Describe a simple algebraic formula.
E=mc2 is the fundamental algebraic formula.
Q2 Are important algebraic formulas and identities 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:
We have learned the important algebraic formulas and identities and their derivation for the ease of students. We have also solved the practice questions related to square, cube, and quadratic formulas. These formulas are essential for a strong grip on algebra.
