Analogy #2

Algebra Level 3

$\displaystyle \begin{cases} x - \sqrt{2401} = 0 \\ y^2 - 2401 = 0 \end{cases}$ Which of the following shows the correct relationship between $x$ and $y$ ?

x > y

x \le y

x < y

x \ge y

x = y

1 solution

Zach Abueg
May 24, 2017

$\displaystyle x = \sqrt{2401}$

$\displaystyle y^2 = 2401 \implies y = \pm \sqrt{2401}$

$\displaystyle \begin{cases} x = y, \ y = + \sqrt{2401} \\ x > y, \ y = - \sqrt{2401} \end{cases}$

Hence, $\displaystyle \boxed{x \geq y}$

@Zach Abueg its 2401...

Ravneet Singh - 4 years ago

Man, I need some sleep. Thanks! I've corrected it.

Zach Abueg - 4 years ago

By the way, you could just say 49 instead of sqrt(2401) just FYI. But still a good answer and solution! Keep it up!

Math Nerd 1729 - 2 years, 6 months ago

@Math Nerd 1729 – I am simply being consistent with the problem.

Zach Abueg - 1 year, 8 months ago

Analogy #2

1 solution

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