Algebraic Expressions
Key Concepts
Solved Examples: 30
Practice Questions: 3
Solved Examples
Question 1: Identify the variable, constant and coefficient in the algebraic expression: 7x + 12
In the expression 7x + 12:
Variable = x
Coefficient of x = 7
Constant = 12
Variable = x
Coefficient of x = 7
Constant = 12
Question 2: Find the terms in the algebraic expression: 5a² − 3a + 8
Terms are the parts of an expression separated by + or − signs.
The terms are: 5a², −3a and 8
The terms are: 5a², −3a and 8
Question 3: Identify the numerical coefficient and variable part of each term in: 3p² + 5p − 11
Terms: 3p², 5p and −11
Parts of each term:
3p² → Coefficient = 3, Variable part = p²
5p → Coefficient = 5, Variable part = p
−11 → Constant term
Answer:
Coefficients = 3 and 5
Constant term = −11
Parts of each term:
3p² → Coefficient = 3, Variable part = p²
5p → Coefficient = 5, Variable part = p
−11 → Constant term
Answer:
Coefficients = 3 and 5
Constant term = −11
Question 4: Identify the variables, constants, coefficients and terms in the expression: 7m²n − 2mn² + 4m − 11
The terms in the expression are:
7m²n, −2mn², 4m and −11
Variables = m and n
Coefficients = 7, −2, 4
Constant = −11
7m²n, −2mn², 4m and −11
Variables = m and n
Coefficients = 7, −2, 4
Constant = −11
Question 5: Identify the type of algebraic expression: 3x³ − 2x² + 5x − 7
The expression 3x³ − 2x² + 5x − 7 has four terms: 3x³, −2x², 5x and −7
An expression with four or more terms is called a polynomial.
Question 6: Find the degree of the algebraic expression: 5x³ + 2x² − 7x + 4
The expression 3x³ − 2x² + 5x − 7 has four terms: 3x³, −2x², 5x and −7
An expression with four or more terms is called a polynomial.
An expression with four or more terms is called a polynomial.
Question 7: Find the degree of the algebraic expression: 4x²y³ + 5xy − 6
For a term containing more than one variable, the degree is the sum of the powers of all variables.
For 4x²y³: Degree = 2 + 3 = 5
For xy: Degree = 1 + 1 = 2
The highest degree among all terms is 5.
Answer: Degree of the expression = 5
For 4x²y³: Degree = 2 + 3 = 5
For xy: Degree = 1 + 1 = 2
The highest degree among all terms is 5.
Answer: Degree of the expression = 5
Question 8: Find the degree of the algebraic expression: 7a⁴ − 3a²b² + 5b³
Find the degree of each term:
7a⁴ → Degree = 4
−3a²b² → Degree = 2 + 2 = 4
5b³ → Degree = 3
The highest degree is 4.
7a⁴ → Degree = 4
−3a²b² → Degree = 2 + 2 = 4
5b³ → Degree = 3
The highest degree is 4.
Question 9: Find the degree of the algebraic term: 9p³q²
The degree of a term is the sum of the powers of all variables.
9p³q²
Degree = 3 + 2 = 5
9p³q²
Degree = 3 + 2 = 5
Question 10: Determine the highest power of each variable in the algebraic expression: 7x⁴y²z + 5xy⁵ − 3z⁶ + 8
For variable x: x⁴ and x
Highest power of x = 4
For variable y: y² and y⁵
Highest power of y = 5
For variable z: z and z⁶
Highest power of z = 6
Highest power of x = 4
For variable y: y² and y⁵
Highest power of y = 5
For variable z: z and z⁶
Highest power of z = 6
Question 11: If A x B = AB + 2A – B then find the value of 4 x 6 + 6 x 4
Substituting the values in A x B = AB + 2A – B
4 x 6 = 4 x 6 + 2 x 4 – 6 = 26
6 x 4 = 6 x 4 + 2 x 6 – 4 = 32
So 4 x 6 + 6 x 4 = 58
4 x 6 = 4 x 6 + 2 x 4 – 6 = 26
6 x 4 = 6 x 4 + 2 x 6 – 4 = 32
So 4 x 6 + 6 x 4 = 58
Question 12: If a × b = a − b + (a/b) then find the value of 12 x 3.
Since a × b = a − b + (a/b),
12 x 3 = 12 – 3 + (12/3) = 13
12 x 3 = 12 – 3 + (12/3) = 13
Question 13: Two numbers a and b (a > b) are such that their sum is equal to five times their difference. Find the value of 4ab / (a² − b²).
By condition (a + b) = 5 x (a – b)
or, a + b = 5a – 5b
or, 4a = 6b
or, 2a = 3b
Now 4ab / (a² − b²).
= 2x3bxb/[(3b/2)² – b²)
= 6b²/[9b²/4 – b²]
= 6b²/(5b²/4)
= 24/5
or, a + b = 5a – 5b
or, 4a = 6b
or, 2a = 3b
Now 4ab / (a² − b²).
= 2x3bxb/[(3b/2)² – b²)
= 6b²/[9b²/4 – b²]
= 6b²/(5b²/4)
= 24/5
Question 14: {1 + 1/x}{1 + 1/(x + 1)}{1 + 1/(x + 2)}{1 + 1/(x + 3)}
1 + 1/x = (x + 1)/x
1 + 1/(x + 1) = (x + 2)/(x + 1)
1 + 1/(x + 2) = (x + 3)/(x + 2)
1 + 1/(x + 3) = (x + 4)/(x + 3)
Therefore,
(1 + 1/x)(1 + 1/(x + 1))(1 + 1/(x + 2))(1 + 1/(x + 3))
= [(x + 1)/x] × [(x + 2)/(x + 1)] × [(x + 3)/(x + 2)] × [(x + 4)/(x + 3)]
Cancelling the common factors,
= (x + 4)/x
= x/x + 4/x
= 1 + 4/x
1 + 1/(x + 1) = (x + 2)/(x + 1)
1 + 1/(x + 2) = (x + 3)/(x + 2)
1 + 1/(x + 3) = (x + 4)/(x + 3)
Therefore,
(1 + 1/x)(1 + 1/(x + 1))(1 + 1/(x + 2))(1 + 1/(x + 3))
= [(x + 1)/x] × [(x + 2)/(x + 1)] × [(x + 3)/(x + 2)] × [(x + 4)/(x + 3)]
Cancelling the common factors,
= (x + 4)/x
= x/x + 4/x
= 1 + 4/x
Question 15: If (2a + b)/(a + 4b) = 3 then find the value of (a + b)/(a + 2b).
Given, (2a + b)/(a + 4b) = 3
⇒ 2a + b = 3(a + 4b)
⇒ 2a + b = 3a + 12b
⇒ a = −11b
Now, (a + b)/(a + 2b)
= (−11b + b)/(−11b + 2b)
= (−10b)/(−9b)
= 10/9
⇒ 2a + b = 3(a + 4b)
⇒ 2a + b = 3a + 12b
⇒ a = −11b
Now, (a + b)/(a + 2b)
= (−11b + b)/(−11b + 2b)
= (−10b)/(−9b)
= 10/9
Question 16: If 1 < x < 2, find the value of √[(x − 1)²] + √[(x − 3)²]
We know: √(a²) = |a|
Therefore,
√[(x − 1)²] + √[(x − 3)²]
= |x − 1| + |x − 3|
Given: 1 < x < 2
For x − 1:
x − 1 is positive.
Therefore,
|x − 1| = x − 1
For x − 3:
x − 3 is negative.
Therefore,
|x − 3| = −(x − 3) = 3 − x
Hence,
= (x − 1) + (3 − x)
= x − 1 + 3 − x
= 2
Answer: 2
Therefore,
√[(x − 1)²] + √[(x − 3)²]
= |x − 1| + |x − 3|
Given: 1 < x < 2
For x − 1:
x − 1 is positive.
Therefore,
|x − 1| = x − 1
For x − 3:
x − 3 is negative.
Therefore,
|x − 3| = −(x − 3) = 3 − x
Hence,
= (x − 1) + (3 − x)
= x − 1 + 3 − x
= 2
Answer: 2
Question 17: If x/y = 3/2 find the value of: (2x + 3y)/(x − y)
Let, x = 3k and y = 2k
Substituting,
(2x + 3y)/(x − y)
= [2(3k) + 3(2k)]/(3k − 2k)
= (6k + 6k)/k
= 12k/k
= 12
Substituting,
(2x + 3y)/(x − y)
= [2(3k) + 3(2k)]/(3k − 2k)
= (6k + 6k)/k
= 12k/k
= 12
Question 18: If x − y = 4 and xy = 21, find the value of: x² + y²
We know, (x − y)² = x² + y² − 2xy
Therefore,
x² + y² = (x − y)² + 2xy
Substituting the values:
= 4² + 2 × 21
= 16 + 42
= 58
Therefore,
x² + y² = (x − y)² + 2xy
Substituting the values:
= 4² + 2 × 21
= 16 + 42
= 58
Question 19: If x + y + z = 12 and xy + yz + zx = 35, find the value of: x² + y² + z²
We know,
(x + y + z)² = x² + y² + z² + 2(xy + yz + zx)
Therefore,
x² + y² + z² = (x + y + z)² − 2(xy + yz + zx)
Substituting the values:
= 12² − 2 × 35
= 144 − 70
= 74
(x + y + z)² = x² + y² + z² + 2(xy + yz + zx)
Therefore,
x² + y² + z² = (x + y + z)² − 2(xy + yz + zx)
Substituting the values:
= 12² − 2 × 35
= 144 − 70
= 74
Question 20: If x + y + z = 9 and xy + yz + zx = 20, find the value of: (x² + y² + z²)/(xy + yz + zx)
We know,
x² + y² + z²
= (x + y + z)² − 2(xy + yz + zx)
= 9² − 2 × 20
= 81 − 40
= 41
Therefore,
(x² + y² + z²)/(xy + yz + zx)
= 41/20
x² + y² + z²
= (x + y + z)² − 2(xy + yz + zx)
= 9² − 2 × 20
= 81 − 40
= 41
Therefore,
(x² + y² + z²)/(xy + yz + zx)
= 41/20
Question 21: If a * b = (a² + b²)/(a + b), find the value of: x * 2x
Using the given operation,
x * 2x
= [x² + (2x)²]/(x + 2x)
= (x² + 4x²)/(3x)
= 5x²/3x
= 5x/3
x * 2x
= [x² + (2x)²]/(x + 2x)
= (x² + 4x²)/(3x)
= 5x²/3x
= 5x/3
Question 22: If a/3 = b/4 = c/7 then what is (a + b + c)/c equal to?
Let, a/3 = b/4 = c/7 = k
Therefore,
a = 3k, b = 4k and c = 7k
Now,
(a + b + c)/c
= (3k + 4k + 7k)/7k
= 14k/7k
= 2
Therefore,
a = 3k, b = 4k and c = 7k
Now,
(a + b + c)/c
= (3k + 4k + 7k)/7k
= 14k/7k
= 2
Question 23: If 2p/(p^2 – 2p + 1) = 1/4, p ≠ 0, then find the value of p + 1/p.
Given,
2p/(p² − 2p + 1) = 1/4
Cross multiplying,
8p = p² − 2p + 1
Rearranging,
p² − 10p + 1 = 0
Since p ≠ 0, divide the equation by p:
p − 10 + 1/p = 0
Therefore,
p + 1/p = 10
2p/(p² − 2p + 1) = 1/4
Cross multiplying,
8p = p² − 2p + 1
Rearranging,
p² − 10p + 1 = 0
Since p ≠ 0, divide the equation by p:
p − 10 + 1/p = 0
Therefore,
p + 1/p = 10
Question 24: If a/(1 – a) + b/(1 – b) + c/(1 – c) = 1, then find the value of 1/(1 – a) + 1/(1 – b) + 1/(1 – c)
We know,
a/(1 − a) = [1 − (1 − a)]/(1 − a)
= 1/(1 − a) − 1
Similarly,
b/(1 − b) = 1/(1 − b) − 1
c/(1 − c) = 1/(1 − c) − 1
Therefore,
a/(1 − a) + b/(1 − b) + c/(1 − c)
= [1/(1 − a) + 1/(1 − b) + 1/(1 − c)] − 3
Given,
[1/(1 − a) + 1/(1 − b) + 1/(1 − c)] − 3 = 1
Therefore,
1/(1 − a) + 1/(1 − b) + 1/(1 − c)
= 1 + 3
= 4
a/(1 − a) = [1 − (1 − a)]/(1 − a)
= 1/(1 − a) − 1
Similarly,
b/(1 − b) = 1/(1 − b) − 1
c/(1 − c) = 1/(1 − c) − 1
Therefore,
a/(1 − a) + b/(1 − b) + c/(1 − c)
= [1/(1 − a) + 1/(1 − b) + 1/(1 − c)] − 3
Given,
[1/(1 − a) + 1/(1 − b) + 1/(1 − c)] − 3 = 1
Therefore,
1/(1 − a) + 1/(1 − b) + 1/(1 − c)
= 1 + 3
= 4
Question 25: If x + 1/x = 5, then find the value of: 2x/(3x² − 5x + 3)
Given,
x + 1/x = 5
Multiplying both sides by x,
x² + 1 = 5x
Now,
3x² − 5x + 3
= 3(x² + 1) − 5x
Substituting x² + 1 = 5x,
= 3(5x) − 5x
= 15x − 5x
= 10x
Therefore,
2x/(3x² − 5x + 3)
= 2x/10x
= 1/5
x + 1/x = 5
Multiplying both sides by x,
x² + 1 = 5x
Now,
3x² − 5x + 3
= 3(x² + 1) − 5x
Substituting x² + 1 = 5x,
= 3(5x) − 5x
= 15x − 5x
= 10x
Therefore,
2x/(3x² − 5x + 3)
= 2x/10x
= 1/5
Question 26: Factorize completely the cyclic algebraic expression: a²(b − c) + b²(c − a) + c²(a − b)
Given a²(b − c) + b²(c − a) + c²(a − b)
Group the terms
= a²b − a²c + b²c − ab² + ac² − bc²
This is a standard alternating cyclic expression. It factors as
= (a − b)(b − c)(a − c)
∴ a²(b − c) + b²(c − a) + c²(a − b) = (a − b)(b − c)(a − c)
Group the terms
= a²b − a²c + b²c − ab² + ac² − bc²
This is a standard alternating cyclic expression. It factors as
= (a − b)(b − c)(a − c)
∴ a²(b − c) + b²(c − a) + c²(a − b) = (a − b)(b − c)(a − c)
Question 27: Factorize the cyclic polynomial expression by grouping common terms: ab(a − b) + bc(b − c) + ca(c − a)
ab(a − b) + bc(b − c) + ca(c − a)
= a²b − ab² + b²c − bc² + ac² − a²c
Group as
= (a²b − a²c) + (b²c − ab²) + (ac² − bc²)
= a²(b − c) + b²(c − a) + c²(a − b)
Now we can use the standard factorization
= (a − b)(b − c)(a − c)
∴ ab(a − b) + bc(b − c) + ca(c − a) = (a − b)(b − c)(a − c)
= a²b − ab² + b²c − bc² + ac² − a²c
Group as
= (a²b − a²c) + (b²c − ab²) + (ac² − bc²)
= a²(b − c) + b²(c − a) + c²(a − b)
Now we can use the standard factorization
= (a − b)(b − c)(a − c)
∴ ab(a − b) + bc(b − c) + ca(c − a) = (a − b)(b − c)(a − c)
Question 28: Factorize the cyclic polynomial expression by grouping common terms: ab(a − b) + bc(b − c) + ca(c − a)
Expand the expression
a²(b − c) + b²(c − a) + c²(a − b)
= a²b − a²c + b²c − ab² + ac² − bc²
Add −abc + abc, which does not change the expression:
= a²b − a²c − abc + ac² − ab² + abc + b²c − bc²
Group the terms:
= (a²b − a²c − abc + ac²) + (−ab² + abc + b²c − bc²)
Factor each group
= a(ab − ac − bc + c²) + b(−ab + ac + bc − c²)
Notice that the second bracket is the negative of the first
= a(ab − ac − bc + c²) − b(ab − ac − bc + c²)
Take the common factor
= (a − b)(ab − ac − bc + c²)
Now factor the expression inside the bracket:
= (a − b)[a(b − c) − c(b − c)]
= (a − b)(b − c)(a − c)
∴ a²(b − c) + b²(c − a) + c²(a − b) = (a − b)(b − c)(a − c)
a²(b − c) + b²(c − a) + c²(a − b)
= a²b − a²c + b²c − ab² + ac² − bc²
Add −abc + abc, which does not change the expression:
= a²b − a²c − abc + ac² − ab² + abc + b²c − bc²
Group the terms:
= (a²b − a²c − abc + ac²) + (−ab² + abc + b²c − bc²)
Factor each group
= a(ab − ac − bc + c²) + b(−ab + ac + bc − c²)
Notice that the second bracket is the negative of the first
= a(ab − ac − bc + c²) − b(ab − ac − bc + c²)
Take the common factor
= (a − b)(ab − ac − bc + c²)
Now factor the expression inside the bracket:
= (a − b)[a(b − c) − c(b − c)]
= (a − b)(b − c)(a − c)
∴ a²(b − c) + b²(c − a) + c²(a − b) = (a − b)(b − c)(a − c)
Question 29: For any real number x ≥ 3, simplify √[x + 6 + 6√(x − 3)]
x + 6 + 6√(x − 3) can be written as
(x − 3) + 9 + 6√(x − 3)
Let √(x − 3) = a
Then x − 3 = a²
So the expression becomes
√(a² + 6a + 9)
= √[(a + 3)²]
Since x ≥ 3, we have a ≥ 0.
∴ √[(a + 3)²] = a + 3
Substituting back
= √(x − 3) + 3
∴ √[x + 6 + 6√(x − 3)] = 3 + √(x − 3)
(x − 3) + 9 + 6√(x − 3)
Let √(x − 3) = a
Then x − 3 = a²
So the expression becomes
√(a² + 6a + 9)
= √[(a + 3)²]
Since x ≥ 3, we have a ≥ 0.
∴ √[(a + 3)²] = a + 3
Substituting back
= √(x − 3) + 3
∴ √[x + 6 + 6√(x − 3)] = 3 + √(x − 3)
Question 30: For any variable x greater than or equal to 1, simplify the nested double radical expression: √[ 2x − 1 + 2√(x² − x) ]
Given √[2x − 1 + 2√(x² − x)], x ≥ 1
Rewrite the expression inside the outer square root
2x − 1 + 2√(x² − x)
= x + (x − 1) + 2√[x(x − 1)]
= (√x + √(x − 1))²
Hence, √[(√x + √(x − 1))²]
= √x + √(x − 1)
Since x ≥ 1, both √x and √(x − 1) are non-negative.
Rewrite the expression inside the outer square root
2x − 1 + 2√(x² − x)
= x + (x − 1) + 2√[x(x − 1)]
= (√x + √(x − 1))²
Hence, √[(√x + √(x − 1))²]
= √x + √(x − 1)
Since x ≥ 1, both √x and √(x − 1) are non-negative.
What are Algebraic Expressions?
An algebraic expression is a mathematical expression made up of variables, constants, and arithmetic operations such as addition (+), subtraction (−), multiplication (×), and division (÷). Unlike an equation, an algebraic expression does not contain an equals (=) sign.
Variables represent unknown or changing values and are usually denoted by letters such as x, y, or z, while constants are fixed numerical values. Algebraic expressions are used to represent mathematical relationships and simplify calculations without knowing the exact value of the variables.
For example, 3x + 5, a² − 4a + 7, 5m, and (x + y) ÷ 2 are all algebraic expressions. These expressions can be evaluated by substituting values for the variables or manipulated using the rules of algebra.
Variables represent unknown or changing values and are usually denoted by letters such as x, y, or z, while constants are fixed numerical values. Algebraic expressions are used to represent mathematical relationships and simplify calculations without knowing the exact value of the variables.
For example, 3x + 5, a² − 4a + 7, 5m, and (x + y) ÷ 2 are all algebraic expressions. These expressions can be evaluated by substituting values for the variables or manipulated using the rules of algebra.
Terms in an Algebraic Expression
A term is a part of an algebraic expression separated by the symbols + or −.
Example: 6x² − 5x + 9
Terms are: 6x², −5x and 9
Each term may contain a variable, a constant, or both.
Example: 6x² − 5x + 9
Terms are: 6x², −5x and 9
Each term may contain a variable, a constant, or both.
Types of Algebraic Expressions
Algebraic expressions are classified according to the number of terms they contain.
Monomial: One term, Example: 8x
Binomial: Two terms, Example: x + 5
Trinomial: Three terms, Example: x² + 4x + 3
Polynomial: Four or more terms, Example: x³ + 2x² − 5x + 7
Monomial: One term, Example: 8x
Binomial: Two terms, Example: x + 5
Trinomial: Three terms, Example: x² + 4x + 3
Polynomial: Four or more terms, Example: x³ + 2x² − 5x + 7
Like and Unlike Terms
Like terms have the same variables raised to the same powers.
Examples of like terms: 3x and 8x, 5a² and −2a²
Unlike terms differ in variables or their powers.
Examples of unlike terms: 3x and 3y, 2a and 2a², 4x² and 4x³
Only like terms can be combined by addition or subtraction.
Examples of like terms: 3x and 8x, 5a² and −2a²
Unlike terms differ in variables or their powers.
Examples of unlike terms: 3x and 3y, 2a and 2a², 4x² and 4x³
Only like terms can be combined by addition or subtraction.
Degree of an Algebraic Expression
The degree of a term is the sum of the powers of its variables. The degree of an algebraic expression is the highest degree among all its terms.
Examples:
5x → Degree = 1
7x² + 3 → Degree = 2
2x³ − x + 5 → Degree = 3
4x²y³ → Degree = 2 + 3 = 5
Examples:
5x → Degree = 1
7x² + 3 → Degree = 2
2x³ − x + 5 → Degree = 3
4x²y³ → Degree = 2 + 3 = 5
Addition and Subtraction of Algebraic Expressions
Only like terms can be added or subtracted.
Example 1: (3x + 5) + (2x − 1)
= 3x + 2x + 5 − 1
= 5x + 4
Example 1: (3x + 5) + (2x − 1)
= 3x + 2x + 5 − 1
= 5x + 4
Multiplication of Algebraic Expressions
Multiply the numerical coefficients and then multiply the variables.
Example 1: 3x × 4 = 12x
Example 2: 2x × 5y = 10xy
Example 3: 3x(2x + 4) = 6x² + 12x
Example 1: 3x × 4 = 12x
Example 2: 2x × 5y = 10xy
Example 3: 3x(2x + 4) = 6x² + 12x
Division of Algebraic Expressions
Divide the numerical coefficients and then divide the variables with the same base.
Example 1: 12x ÷ 3 = 4x
Example 2: 20x² ÷ 5x = 4x
Example 1: 12x ÷ 3 = 4x
Example 2: 20x² ÷ 5x = 4x
Summary of Algebraic Expressions
| Concept | Summary | Example |
|---|---|---|
| Algebraic Expression | A mathematical expression containing variables, constants and arithmetic operations, but no equals (=) sign. | 3x + 5 |
| Variable | A symbol representing an unknown or changing value. | x in 4x + 7 |
| Constant | A fixed numerical value. | 7 in 4x + 7 |
| Coefficient | The numerical factor multiplying a variable. | 4 in 4x + 7 |
| Term | A part of an algebraic expression separated by + or − signs. | 6x², −5x and 9 in 6x² − 5x + 9 |
| Types of Expressions | Expressions classified based on the number of terms they contain. | Monomial: 5x Binomial: x + 2 Trinomial: x² + x + 1 |
| Like Terms | Terms having the same variables raised to the same powers. | 3x and 8x |
| Unlike Terms | Terms having different variables or different powers. | 2x and 2x² |
| Degree | The highest power of the variable in an algebraic expression. | Degree of 3x³ − 2x + 5 is 3 |
| Addition & Subtraction | Only like terms can be added or subtracted. | 3x + 2x = 5x |
| Multiplication | Multiply coefficients and variables separately. | 2x × 5y = 10xy |
| Division | Divide coefficients and variables with the same base. | 20x² ÷ 5x = 4x |
| Simplification | Combining like terms and performing operations to write an expression in its simplest form. | 5x + 2 − 3x + 6 = 2x + 8 |
Common Mistakes
- Confusing terms and factors: Terms are separated by + or − signs, while factors are quantities multiplied within a term.
- Combining unlike terms: Only terms with the same variables having the same powers can be added or subtracted.
- Ignoring negative signs: The sign before a term is part of the term and must be considered while simplifying expressions.
- Finding degree incorrectly: For terms with multiple variables, the degree is the sum of the powers of all variables.
- Making substitution errors: Always substitute the given values carefully using brackets, especially when the value is negative.
- Cancelling incorrectly in algebraic fractions: Only common factors can be cancelled; terms connected by addition or subtraction cannot be cancelled.
Practice Questions
Question 1: If a + b = 15 and ab = 44, find the value of: (a² + b²)/(a² − 2ab + b²)
Question 2 : The sum of two numbers is 18 and their product is 77. Find the value of: (Larger number)² + (Smaller number)²
Question 3: The length and breadth of a rectangular field are represented by (x + 3) m and (x − 3) m respectively. If the area of the field is 91 m², find the value of: x² + 9/x²
Related Topics
 |
|
|
|
|
|