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

From Application Problems on Expansion of Powers of Binomials and Trinomials to HOME PAGE