Question
Easy

If $x^2 + \left( \frac{ax}{a+x} \right)^2 = 3a^2 ; x \ne$ -a then one of the value of x is equal to :

1
$\frac{1+\sqrt{5}}{2}$
2
$\frac{a+a\sqrt{5}}{2}$
3
$\frac{1+\sqrt{3}}{2}$
4
$\frac{a+a\sqrt{3}}{2}$
Question Details
Time to Solve: 12
Exam: HTET
Level/Paper: Level 2
Chapter: Algebra
Topic: Quadratic Equations
Correct Answer
Option B
Explanation

To solve the given problem and justify why Option 2 is the correct answer, let's analyze the equation and the options provided. The equation given is: \[ x^2 + \left( \frac{ax}{a+x} \right)^2 = 3a^2 \] We need to find a value of \( x \) that satisfies this equation, given that \( x \neq -a \). ### Step-by-Step Explanation: 1. Substitute Option 2 into the Equation: Option 2 suggests that…Read More

If x2 left - HTET Level 2 | Clear Cutoff