Question

Show that the given square is a magic square (i.e. all of the rows, columns and diagonals have the same sum).

$6 a$	$7 a^{2} - 11 a + 24$	$- a^{2} - 7 a - 6$
$a^{2} - 17 a$	$2 (a^{2} - 2 a + 3)$	$- 3 (- a^{2} - 3 a - 4)$
$5 a^{2} - a + 18$	$- 3 (a^{2} + 4 - a)$	$2 (2 a^{2} - 7 a + 6)$

Accepted Answer

Given square:

$6 a$ $7 a^{2} - 11 a + 24$ $- a^{2} - 7 a - 6$

$a^{2} - 17 a$ $2 (a^{2} - 2 a + 3)$ $- 3 (- a^{2} - 3 a - 4)$

$5 a^{2} - a + 18$ $- 3 (a^{2} + 4 - a)$ $2 (2 a^{2} - 7 a + 6)$

To show this square is a magic square we have to do the sum of the numbers in each row, column and diagonal and if it is the same then this square is a magic square.

First, we sum all the rows

Row1: $s u m = 6 a + 7 a^{2} - 11 a + 24 - a^{2} - 7 a - 6 = 6 a^{2} - 12 a + 18$
Row2: $s u m = a^{2} - 17 a + 2 (a^{2} - 2 a + 3) - 3 (- a^{2} - 3 a - 4) = a^{2} - 17 a + 2 a^{2} - 4 a + 6 + 3 a^{2} + 9 a + 12 = 6 a^{2} - 12 a + 18$
Row3: $s u m = 5 a^{2} - a + 18 - 3 (a^{2} + 4 - a) + 2 (2 a^{2} - 7 a + 6) = 5 a^{2} - a + 18 - 3 a^{2} - 12 + 3 a + 4 a^{2} - 14 a + 12 = 6 a^{2} - 12 a + 18$
Now, we sum all the columns

Column1: $s u m = 6 a + a^{2} - 17 a + 5 a^{2} - a + 18 = 6 a^{2} - 12 a + 18$
Column2: $s u m = 7 a^{2} - 11 a + 24 + 2 (a^{2} - 2 a + 3) - 3 (a^{2} + 4 - a) = 7 a^{2} - 11 a + 24 + 2 a^{2} - 4 a + 6 - 3 a^{2} - 12 + 3 a = 6 a^{2} - 12 a + 18$
Column3: $s u m = - a^{2} - 7 a - 6 - 3 (- a^{2} - 3 a - 4) + 2 (2 a^{2} - 7 a + 6) = - a^{2} - 7 a - 6 + 3 a^{2} + 9 a + 12 + 4 a^{2} - 14 a + 12 = 6 a^{2} - 12 a + 18$
Now, we sum both the diagonals

Diagonal1: $s u m = 6 a + 2 (a^{2} - 2 a + 3) + 2 (2 a^{2} - 7 a + 6) = 6 a + 2 a^{2} - 4 a + 6 + 4 a^{2} - 14 a + 12 = 6 a^{2} - 12 a + 18$
Diagonal2: $s u m = - a^{2} - 7 a - 6 + 2 (a^{2} - 2 a + 3) + 5 a^{2} - a + 18 = - a^{2} - 7 a - 6 + 2 a^{2} - 4 a + 6 + 5 a^{2} - a + 18 = 6 a^{2} - 12 a + 18$
Here the sum of the number in each row, column and diagonal is the same. That means it is a 'MAGIC SQUARE'.

$2 (a + b)$	$3 a + b$	$a - 12 b$
$a - 17 b$	$2 a - 3 b$	$3 a + 11 b$
$- 3 (- a - 2 b)$	$a - 7 b$	$2 (a - 4 b)$

$3 (a - b)$	$- 9 a + 12 b$	$3 b$
$- 5 (a - 2 b)$	$2 (- a + 2 b)$	$- (2 b - a)$
$- 4 a + 5 b$	$5 a - 4 b$	$- 7 a + 11 b$

$- 23 a + 16 b$	$5 (3 a - 2 b)$	$- 19 a + 3 b$
$5 (- a - 2 b)$	$- 3 (3 a - b)$	$- 13 a + 16 b$
$a + 3 b$	$- 33 a + 16 b$	$5 (a - 2 b)$

David Weber, Talei Kunkel, Alexander Martinez and, Rebecca Shultis Solutions for Chapter: Algebraic Expressions and Equations Puzzles and Tricks, Exercise 10: Practice 3

David Weber Mathematics Solutions for Exercise - David Weber, Talei Kunkel, Alexander Martinez and, Rebecca Shultis Solutions for Chapter: Algebraic Expressions and Equations Puzzles and Tricks, Exercise 10: Practice 3

Questions from David Weber, Talei Kunkel, Alexander Martinez and, Rebecca Shultis Solutions for Chapter: Algebraic Expressions and Equations Puzzles and Tricks, Exercise 10: Practice 3 with Hints & Solutions

Learn and Prepare for any exam you want !


$- 3 m^{3} + 5 m^{2}$	$- 2 (m^{3} - 2 m^{2})$
$2 (2 m^{3} - 3 m^{2})$

		$- 4 t + 5 u$
		$- 2 (- 3 t - 2 u)$
		$- 5 (t - 3 u)$