Question
Easy
If ($x^4 + \frac{1}{x^4})$ = 119 and x > 1, then the value of ($x^3 - \frac{1}{x^3}$) is:
1
18
2
36
3
54
4
72
Question Details
Time to Solve: 12
Exam: HTET
Level/Paper: Level 2
Chapter: Power & Roots
Topic: Exponents & Powers
Correct Answer
Option B
Explanation
To solve the problem and justify why Option 2 is correct, we need to find the value of \(x^3 - \frac{1}{x^3}\) given that \(x^4 + \frac{1}{x^4} = 119\). ### Step-by-step Explanation: 1. Given Equation: \[ x^4 + \frac{1}{x^4} = 119 \] 2. Relate to \(x^2 + \frac{1}{x^2}\): We know that: \[ \left(x^2 + \frac{1}{x^2}\right)^2 = x^4 + 2 + \frac{1}{x^4} \] Therefore: \[ x^4 + \frac{1}{x^4} = \left(x^2 + \frac{1}{x^2}\right)^2…Read More
Similar Questions from REET Exam - Paper 1 - Year 2018
Question 1
Easy
Source :
HTET 2023
If ($x + \frac{1}{x}$) = 2, then the value of ($\sqrt{x} + \frac{1}{\sqrt{x}}$) is :
Chapter :
Algebra
Topic :
Equations
Question 2
Easy
Source :
HTET 2023
If the sum of n terms of an arithmetic progression is (3n$^2$ + 5n), then 158 is :
Chapter :
Algebra
Topic :
Sequences and Series (AP, GP, HP)
Question 3
Easy
Source :
HTET 2023
If S is the incentre of triangle ABC and ∠A = 30°, then the value of ∠BSC is:
Chapter :
2D Geometry
Topic :
Triangles
Question 4
Easy
Source :
HTET 2023
If a = $\frac{\sqrt5+1}{\sqrt5-1}$ and b = $\frac{\sqrt5-1}{\sqrt5+1}$, then the value of $\frac{a^2 + ab + b^2}{a^2 - ab + b^2}$ is:
Chapter :
Power & Roots
Topic :
Surds & Indices