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• Module 4 Day 6 Challenge Part 1

  Nested Square Root, Expanding (√x + √y)²

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  @The-Blade-Dancer Thank you for writing and for being willing to share your comments. 🙂 Looking back on this explanation, I sort of want to smack my head 🤦 ... You're right, it is a bit obtuse. (Okay not a bit, but very!)

  temp-20201024-4.png

  The question is asking whether \( a + b\sqrt{c} = x + y + 2\sqrt{xy} \) for integers \(a, b,\) and \(c\) implies that \(x\) and \(y\) are always rational. (Rational means that the number is of the form \(\frac{m}{n}\) where \(m\) and \(n\) are both integers.) Hmmm... it seems like this statement can't be true, just from using some common sense, because how can we be so lucky as to always get \(x\) and \(y\) rational... but how do we prove this?

  Luckily, with a question of the form "It's always true that.... ", all we have to do is find a single counterexample where it isn't true.

  I've updated the explanation to hopefully be a bit more simple and easy to understand. The counterexample I'm using is a case where \(x\) and \(y\) are both imaginary numbers but \(a,\) \(b,\) and \(c\) are integers.

  temp-20201024-3.png

• Module 4 Day 6 Challenge Part 2

  System of 2 Equations in 2 Variables, Square Roots

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  Great question, @the-blade-dancer !

  If you remember the binomial theorem at all from Module 3 (let me know if you need more explanation here), but we can use a shortcut for the expansion of \((\sqrt{a}+\sqrt{b})^3\).

  In general, \((x+y)^3=x^3+3x^2y+3xy^2+y^3\), so

  \((\sqrt{a}+\sqrt{b})^3=a\sqrt a +3a\sqrt b+ 3\sqrt a b+b\sqrt b\)
  \((\sqrt{a}+\sqrt{b})^3= (a+3b)\sqrt a+(3a+b)\sqrt b\)

  This gets us straight to the form that we need. And since this has to be equal to \(22\sqrt 7+26\sqrt 5\), it is most likely that \(a=7, b=5\), so we can go and check that out. After confirming that it is true, we get that our answer is \(7+5=\boxed{12}\). Let me know if you have any other questions!

• Module 4 Day 6 Challenge Part 3

  Simplifying Rational Radical Expressions, Rationalizing the Denominator, Difference of Squares

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  @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.

  🙂

• Module 4 Day 6 Your Turn Part 1

  Integer Equations Relating Sum and Product of Two Unknowns

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• Module 4 Day 6 Your Turn Part 2

  Shortcut Using Reciprocals

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