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.
From Application Problems on Expansion of Powers of Binomials and Trinomials to HOME PAGE
Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.
Dec 11, 24 09:08 AM
Dec 09, 24 10:39 PM
Dec 09, 24 01:08 AM
Dec 08, 24 11:19 PM
Dec 07, 24 03:38 PM
New! Comments
Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.