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.