Application Problems on Expansion of Powers of Binomials and Trinomials are available here. So, the students who are in search of problems related to the expansion of powers of binomials and trinomials can use this page and practice the sums. In this article, you can find different types of problems on the expansion of powers of binomials and trinomials with step-by-step explanations. Practice the given questions and test your knowledge on this topic.
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Application Problems on Expansion of Powers of Binomials and Trinomials
Students of 9th grade can make use of the below examples and improve their math skills.
Example 1.
Use (x ± y)² = x² ± 2xy + y² to evaluate (3.07)²
Solution:
(3.07)²
We can evaluate (3.07)² by using (x + y)² = x² + 2xy + y²
(3 + 0.07)²
3² + 2 × (3)(0.07) + (0.07)²
9 + 2 × 0.21 + 0.0049
9 + 0.42 + 0.0049
9.4249
Example 2.
Use (x ± y)² = x² ± 2xy + y² to evaluate (8.92)².
Solution:
(8.92)²
We can evaluate (8.92)² by using (x – y)² = x² – 2xy + y²
(9 – 0.08)²
9² – 2(9)(0.08) + (0.08)²
81 – 1.44 + 0.0064
81 – 1.4464
79.553
Example 3.
Evaluate 24 × 16 using (x + y)(x – y) = x² – y²
Solution:
(20 + 4)(20 – 4)
20² – 4²
400 – 16
384
Example 4.
Evaluate 6.98 × 7.02
Solution:
We can evaluate 6.98 × 7.02 by using the formula (x + y)(x – y) = x² – y²
(7 – 0.02)(7 + 0.02)
7² – 0.02²
49 – 0.0004
48.99
Example 5.
If the sum of two numbers x and y is 20 and the sum of their squares is 56 find the product of the numbers.
Solution:
Sum of two numbers x and y is 20
Sum of the two numbers x and y squares is 56
x² + y² = 56
We know that
2ab = (a + b)² – (a² + b²)
2xy = 20² – 56
200 – 56
2xy = 144
Therefore xy = 1/2 × 2xy
1/2 × 144 = 72.
Example 6.
If the sum of three numbers a, b, c is 3 and the sum of their squares is 12 then find the sum of the products of the three numbers taking two at a time.
Solution:
Sum of three numbers a, b, c is 3
a + b + c = 3
Sum of three numbers a, b, c squares is 12
a² + b² + c² = 12
We need to find the value of ab + bc + ca
We know that
(a + b + c)² = a² + b² + c² + 2(ab + bc + ca)
(a + b + c)² – a² + b² + c² = 2(ab + bc + ca)
3² – 12 = 2( ab + bc + ca)
9 – 12 = 2( ab + bc + ca)
-3 = 2(ab + bc + ca)
ab + bc + ca = -2/3
Example 7.
Evaluate (2.34)² + (3.14)²
Solution:
a³ + b³ = (a + b)³ – 3ab(a + b)
(2.34)³ + (3.14)³
(2.34 + 3.14)³ – 3(2.34)(3.14)(2.34+3.14)
164.56 – 120.79
43.77
Example 8.
If the sum of two numbers is 6 and the sum of their cubes is 27, find the sum of their squares
Solution:
Given,
Sum of the numbers is 6
The two numbers are a, b.
a + b = 6
Sum of their cubes is 27
a³ + b³ = 27
a³ + b³ = (a + b)³ – 3ab(a + b)
6³ – 27 = 3ab(a + b)
18ab = 216 – 27 = 189
ab = 189/18 = 23.625
Now a² + b² = (a + b)² – 2ab
6² – 2 × 23. 625
36 – 2 × 23.625
803.26
Example 9.
Use (x ± y)² = x² ± 2xy + y² to evaluate (1.07)²
Solution:
Given that,
(1.07)²
(1 + 0.07)²
1² + 2 × (1)(0.07) + (0.07)²
1 + 2 × 0.07 + 0.0049
1 + 0.14 + 0.0049
1.1449
Example 10.
If the sum of two numbers x and y is 5 and the sum of their squares is 12 find the product of the numbers.
Solution:
Sum of two numbers x and y is 5
Sum of the two numbers x and y squares is 12
x² + y² = 12
We know that
2ab = (a + b)² – (a² + b²)
2xy = 5² – 12
25 – 12
2xy = 13
Therefore xy = 1/2 × 2xy
1/2 × 13 = 13/2.