# 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.

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