Forum — Daily Challenge
    • Categories
    • Recent
    • Tags
    • Popular
    • Users
    • Groups
    • Login

    Why does choice i. in the mini question not work? I think it's fine.

    Module 4 Day 6 Challenge Part 3
    2
    2
    50
    Loading More Posts
    • Oldest to Newest
    • Newest to Oldest
    • Most Votes
    Reply
    • Reply as topic
    Log in to reply
    This topic has been deleted. Only users with topic management privileges can see it.
    • The Blade DancerT
      The Blade Dancer M0★ M1★ M2★ M3★ M4 M5
      last edited by debbie

      Module 4 Week 2 Day 6 Challenge Part 3 Mini-Question

      Why does choice i. in the mini question not work? I think it's fine.

      The Blade Dancer
      League of Legends, Valorant: Harlem Charades (#NA1)
      Discord: Change nickname if gay#7585

      debbieD 1 Reply Last reply Reply Quote 2
      • debbieD
        debbie ADMIN M0★ M1 M5 @The Blade Dancer
        last edited by

        @The-Blade-Dancer Thank you for asking and thanks for being curious. 🙂

        The reason choice i.) is incorrect is because in general, for any numbers \(a\) and \(b,\) the following isn't true:

        $$ \sqrt{ a + b } = \sqrt{a} + \sqrt{b} $$

        You could try some values of \(a\) and \(b\) to see this. Like for example, \(a = 1\) and \(b = 4.\) Then let's check our statement.

        $$ \sqrt{ 1 + 4 } \stackrel{?}{=} \sqrt{1} + \sqrt{4} $$

        $$ \sqrt{ 5} \stackrel{?}{=} 1 + 2 $$
        $$ \sqrt{5} \neq 3 $$

        The answer choice i.) is claiming that the following is true:

        $$ \sqrt{ 6 + 5\sqrt{2}} = \sqrt{6} + \sqrt{5\sqrt{2}} $$

        However, this is just the same as the statement \(\sqrt{a + b} = \sqrt{a} + \sqrt{b} \) where \(a = 6\) and \(b = 5\sqrt{2}.\) And since we just showed that this statement is false, then choice i.) can't be true.

        🙂

        1 Reply Last reply Reply Quote 2

        • 1 / 1
        • First post
          Last post
        Daily Challenge | Terms | COPPA