# Khan/khan-exercises

Fix #33076 remove extraneous hint

Summary: There's no reason for one-root to inherit from original

Test Plan: Tested locally

Reviewers: stephanie

Reviewed By: stephanie

Differential Revision: http://phabricator.khanacademy.org/D897
1 parent d3ce772 commit c9bb36efa9890c1a0b79c667c35b190ff55211fc beneater committed Nov 13, 2012
74 exercises/simplifying_square_roots_of_negatives.html
 @@ -0,0 +1,74 @@ + + + + + Simplifying square roots of negative numbers + + + +
+
+
+
+ randFromArray([30, 42, 66, 70, 105, 110, 154, 165, 210]) +
+ +

+ Simplify \sqrt{-NUM}. +

+ +
NUM
+ +
+

The number NUM has no perfect square factors.

+

Remember that \sqrt{-1} = i.

+

Thus, \sqrt{-NUM} = \sqrt{NUM \cdot (-1)} = \sqrt{NUM}~i is the simplest form

+
+
+ +
+
+ randFromArray([9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225]) + splitRadical(NUM) + SPLIT[0] === 1 ? "" : SPLIT[0] +
+ +

+ Simplify \sqrt{-NUM}. +

+ +
NUM
+ +
+

The number NUM is a perfect square, so -NUM = COEFFICIENT^2 \cdot (-1).

+

Remember that \sqrt{-1} = i.

+

Thus, \sqrt{-NUM} = \sqrt{COEFFICIENT^2 \cdot (-1)} = COEFFICIENTi.

+
+
+ +
+
+ randFromArray([8, 12, 18, 20, 24, 27, 28, 32, 40, 44, 45, 48, 50, 54, 56, 60, 63, 72, 75, 80, 88, 90, 98, 99, 120, 125, 128, 140, 150, 160, 175, 180, 200, 216]) + splitRadical(NUM) + SPLIT[0] === 1 ? "" : SPLIT[0] + SPLIT[1] === 1 ? "" : SPLIT[1] +
+ +

+ Simplify \sqrt{-NUM}. +

+ +
NUM
+ +
+

The largest perfect square that divides NUM is COEFFICIENT * COEFFICIENT.

+

Factoring it out, we have NUM = COEFFICIENT^2 \cdot RADICAL.

+

So \sqrt{-NUM} = \sqrt{COEFFICIENT^2 \cdot RADICAL \cdot (-1)}.

+

Remember that \sqrt{-1} = i.

+

+
+
+
+
+ +
 @@ -41,7 +41,7 @@
-
+

The two numbers -A and -B satisfy both conditions:

\qquad \color{PINK}{-A} + \color{PINK}{-B} = @@ -52,22 +52,22 @@ \color{BLUE}{SIMPLECONSTANT}

-

+

So (x A < 0 ? "+" : "" \color{PINK}{-A}) (x B < 0 ? "+" : "" \color{PINK}{-B}) = 0.

-

+

Since (x A < 0 ? "+" : "" -A) (x B < 0 ? "+" : "" -B) = 0, we know that one or both quantities must equal zero for the equation to be true.

-

x + -A = 0 or x + -B = 0

-

Thus, x = A and x = B are the solutions.

+

x + -A = 0 or x + -B = 0

+

Thus, x = A and x = B are the solutions.

-
+
SQUARE * A * A A * A @@ -80,7 +80,7 @@

-
+

The number -A used twice satisfies both conditions:

\qquad \color{PINK}{-A} + \color{PINK}{-A} = @@ -91,9 +91,9 @@ \color{BLUE}{SIMPLECONSTANT}

-

So (x A < 0 ? "+" : "" \color{PINK}{-A})^2 = 0.

-

x + -A = 0

-

Thus, x = A is the solution.

+

So (x A < 0 ? "+" : "" \color{PINK}{-A})^2 = 0.

+

x + -A = 0

+

Thus, x = A is the solution.

1 test/exercises/simplifying_square_roots_of_negatives.js
 @@ -444,7 +444,10 @@ $.extend(Khan.answerTypes, { solutionarea.addClass("radical") .append(inte) .append('') - .append(rad.addClass("overline")); + .append(rad.addClass("overline")) + if (options.complexRadical !== undefined) { + solutionarea.append('i') + }; var ret = function() { // Load entered values into inteGuess, radGuess @@ -470,7 +473,11 @@$.extend(Khan.answerTypes, { } }; if (options.simplify === "required") { - ret.examples = ["a simplified radical, like \\sqrt{2} or 3\\sqrt{5}"]; + if (options.complexRadical !== undefined) { + ret.examples = ["a simplified radical, like \\sqrt{2}~i or 3\\sqrt{5}~i"]; + } else { + ret.examples = ["a simplified radical, like \\sqrt{2} or 3\\sqrt{5}"]; + } } else { ret.examples = ["a radical, like \\sqrt{8} or 2\\sqrt{2}"]; }