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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 = x2 ± 2xy + y2 to evaluate (2.05)2.

Solution:

(2.05)2

= (2 + 0.05)2

= 22 + 2 × 2 × 0.05 + (0.05)2

= 4 + 0.20 + 0.0025

= 4.2025.

2. Use (x ± y)2 = x2 ± 2xy + y2 to evaluate (5.94)2.

Solution:

(5.94)2

= (6 – 0.06)2

= 62 – 2 × 6 × 0.06 + (0.06)2

= 36 – 0.72 + 0.0036

= 36.7236.


3. Evaluate 149 × 151 using (x + y)(x - y) = x2 - y2

Solution:

149 × 151

= (150 - 1)(150 + 1)

= 1502 - 12

= 22500 - 1

= 22499


4. Evaluate 3.99 × 4.01 using (x + y)(x - y) = x2 - y2.

Solution:

3.99 × 4.01

= (4 – 0.01)(4 + 0.01)

= 42 - (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., x2 + y2 = 52

We know that, 2ab = (a + b)2 – (a2 + b2)

Therefore, 2xy = (x + y)2 - (x2 + y2)

           ⟹ 2xy = 102 - 52

           ⟹ 2xy = 100 - 52

           ⟹ 2xy = 48

Therefore, xy = 12 × 2xy

                    = 12 × 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., p2 + q2+ r2= 14

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

We know that, (a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca).

Therefore, (p + q + r)2 = p2 + q2 + r2 + 2(pq + qr + rp).

⟹ (p + q + r)2 - (p2 + q2 + r2) = 2(pq + qr + rp).

⟹ 62 - 14 = 2(pq + qr + rp).

⟹ 36 – 14 = 2(pq + qr + rp).

⟹ 22 = 2(pq + qr + rp).

⟹ pq + qr + rp = 222

Therefore, pq + qr + rp = 11.


7. Evaluate: (3.29)3 + (6.71)3

Solution:

We know, a3 + b3 = (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)

= 103 – 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., a3 + b3 = 189

Now a3 + b3 = (a + b) 3 – 3ab(a + b).

Therefore, 93 – 189 = 3ab × 9.

Therefore, 27ab = 729 – 189 = 540.

Therefore, ab = 54027 = 20.

Now, a2 + b2 = (a + b)2 – 2ab

                                           = 92 – 2 × 20

                                           = 81 – 40

                                           = 41.

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





9th Grade Math

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