Application Problems on Expansion of Powers of Binomials and Trinomials

Here we will solve different types of application problems on expansion of powers of binomials and trinomials.

1. Use (x ± y)\(^{2}\) = x\(^{2}\) ± 2xy + y\(^{2}\) to evaluate (2.05)\(^{2}\).

Solution:

(2.05)\(^{2}\)

= (2 + 0.05)\(^{2}\)

= 2\(^{2}\) + 2 × 2 × 0.05 + (0.05)\(^{2}\)

= 4 + 0.20 + 0.0025

= 4.2025.

2. Use (x ± y)\(^{2}\) = x\(^{2}\) ± 2xy + y\(^{2}\) to evaluate (5.94)\(^{2}\).

Solution:

(5.94)\(^{2}\)

= (6 – 0.06)\(^{2}\)

= 6\(^{2}\) – 2 × 6 × 0.06 + (0.06)\(^{2}\)

= 36 – 0.72 + 0.0036

= 36.7236.


3. Evaluate 149 × 151 using (x + y)(x - y) = x\(^{2}\) - y\(^{2}\)

Solution:

149 × 151

= (150 - 1)(150 + 1)

= 150\(^{2}\) - 1\(^{2}\)

= 22500 - 1

= 22499


4. Evaluate 3.99 × 4.01 using (x + y)(x - y) = x\(^{2}\) - y\(^{2}\).

Solution:

3.99 × 4.01

= (4 – 0.01)(4 + 0.01)

= 4\(^{2}\) - (0.01)\(^{2}\)

= 16 - 0.0001

= 15.9999


5. If the sum of two numbers x and y is 10 and the sum of their squares is 52, find the product of the numbers.

Solution:

According to the problem, sum of two numbers x and y is 10

i.e., x + y = 10 and

Sum of the two numbers x and y squares is 52

i.e., x\(^{2}\) + y\(^{2}\) = 52

We know that, 2ab = (a + b)\(^{2}\) – (a\(^{2}\) + b\(^{2}\))

Therefore, 2xy = (x + y)\(^{2}\) - (x\(^{2}\) + y\(^{2}\))

           ⟹ 2xy = 10\(^{2}\) - 52

           ⟹ 2xy = 100 - 52

           ⟹ 2xy = 48

Therefore, xy = \(\frac{1}{2}\) × 2xy

                    = \(\frac{1}{2}\) × 48

                    = 24.


6. If the sum of three numbers p, q, r is 6 and the sum of their squares is 14 then find the sum of the products of the three numbers taking two at a time.

Solution:

According to the problem, sum of three numbers p, q, r is 6.

i.e., p + q + r = 6 and

Sum of the three numbers p, q, r squares is 14

i.e., p\(^{2}\) + q\(^{2}\)+ r\(^{2}\)= 14

Here we need to find the value of pq + qr + rp

We know that, (a + b + c)\(^{2}\) = a\(^{2}\) + b\(^{2}\) + c\(^{2}\) + 2(ab + bc + ca).

Therefore, (p + q + r)\(^{2}\) = p\(^{2}\) + q\(^{2}\) + r\(^{2}\) + 2(pq + qr + rp).

⟹ (p + q + r)\(^{2}\) - (p\(^{2}\) + q\(^{2}\) + r\(^{2}\)) = 2(pq + qr + rp).

⟹ 6\(^{2}\) - 14 = 2(pq + qr + rp).

⟹ 36 – 14 = 2(pq + qr + rp).

⟹ 22 = 2(pq + qr + rp).

⟹ pq + qr + rp = \(\frac{22}{2}\)

Therefore, pq + qr + rp = 11.


7. Evaluate: (3.29)\(^{3}\) + (6.71)\(^{3}\)

Solution:

We know, a\(^{3}\) + b\(^{3}\) = (a + b) \(^{3}\) – 3ab(a + b)

Therefore, (3.29)\(^{3}\) + (6.71)\(^{3}\)

= (3.29 + 6.71)\(^{3}\) – 3 × 3.29 × 6.71(3.29 + 6.71)

= 10\(^{3}\) – 3 × 3.29 × 6.71 × 10

= 1000 - 3 × 220.759

= 1000 – 662.277

= 337.723


14. If the sum of two numbers is 9 and the sum of their cubes is 189, find the sum of their squares.

Solution:

Let a, b are the two numbers

According to the problem, sum of two numbers is 9

 i.e., a + b = 9 and

Sum of their cubes is 189

i.e., a\(^{3}\) + b\(^{3}\) = 189

Now a\(^{3}\) + b\(^{3}\) = (a + b) \(^{3}\) – 3ab(a + b).

Therefore, 9\(^{3}\) – 189 = 3ab × 9.

Therefore, 27ab = 729 – 189 = 540.

Therefore, ab = \(\frac{540}{27}\) = 20.

Now, a\(^{2}\) + b\(^{2}\) = (a + b)\(^{2}\) – 2ab

                                           = 9\(^{2}\) – 2 × 20

                                           = 81 – 40

                                           = 41.

Therefore, the sum of the squares of the numbers is 41.





9th Grade Math

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