>
Exams
>
Quantitative Aptitude
>
Geometric Progressions
>
1 sqrt 2 4 1 sqrt 2 4
Question:
\[ (1 + \sqrt{2})^4 + (1 - \sqrt{2})^4 \]
Show Hint
Use the binomial expansion to simplify powers of binomials and combine like terms.
AP ICET - 2024
AP ICET
Updated On:
Apr 28, 2025
34
17
51
68
Hide Solution
Verified By Collegedunia
The Correct Option is
A
Solution and Explanation
We can expand the terms \( (1 + \sqrt{2})^4 \) and \( (1 - \sqrt{2})^4 \) using the binomial expansion: \[ (1 + \sqrt{2})^4 = 1 + 4\sqrt{2} + 6 \times 2 + 4\sqrt{2} + 4 = 17 + 8\sqrt{2}, \] and \[ (1 - \sqrt{2})^4 = 1 - 4\sqrt{2} + 6 \times 2 - 4\sqrt{2} + 4 = 17 - 8\sqrt{2}. \] Now, adding both expressions: \[ (1 + \sqrt{2})^4 + (1 - \sqrt{2})^4 = (17 + 8\sqrt{2}) + (17 - 8\sqrt{2}) = 3(4) \]
Download Solution in PDF
Was this answer helpful?
0
0
Top Questions on Geometric Progressions
If \( a, b, c \) are in Geometric Progression and \( a^x = b^y = c^z \), then \( x, y, z \) are in:
CUET (PG) - 2025
Computer Science
Geometric Progressions
View Solution
Two numbers are in the ratio of 4 : 7. If 14 is added to each, they are in the ratio 5 : 7, then find the numbers?
AP ICET - 2024
Quantitative Aptitude
Geometric Progressions
View Solution
What is the Geometric Mean of 12, 30, and 75?
AP ICET - 2024
Quantitative Aptitude
Geometric Progressions
View Solution
If the 6th term of a Geometric Progression (GP) is 243 and the 1st term is 32, then what will be the 5th term of the GP?
AP POLYCET - 2020
Mathematics
Geometric Progressions
View Solution
View All
Questions Asked in AP ICET exam
If 'ACID' = 1C3D, 'PAMPER' = P1MP2R, 'BOMBAY' = B4MB1Y, then 'UNIVERSITY' = ..................................?
AP ICET - 2024
Coding Decoding
View Solution
Find the odd thing from the following:
AP ICET - 2024
Odd one Out
View Solution
Find the odd thing from the following:
AP ICET - 2024
Odd one Out
View Solution
Find the odd thing from the following:
AP ICET - 2024
Odd one Out
View Solution
Find the odd thing from the following:
AP ICET - 2024
Odd one Out
View Solution
View More Questions