You explain the beauty in mathematics elegantly. As a student I didn’t understand that complex numbers are crucial in solving real life problems in physics and engineering. As a retired engineer mathematics still brings me joy.
x must be < 0. Quick testing yields -3 (-1 and -2 can be excluded at hand and -4 is already to large). Polynomial division by (x+3) yields x² - 4x + 12 which yields two complex solutions 2 +/- 2i*SQRT(2).
Even though I taught math, I don't consider myself a math genius. But -3 just jumped out at me. So even if there is an elaborate solution that nobody ever uses, and few could understand, the -3 is THE answer that applies in the everyday world where we all live.
X needs to be negative so that squaring it becomes positive and cubing becomes negative. That way 'subtracting the negative' becomes 'adding a positive'. Then just plugged in integers until reaching -3.
x² - x³ = 36 Looking at what any real answers must be, for positive mumbers x² and x³ are both positive, between -1and 1, x² and x³ are also between -1 and 1, and above 1, x³ > x², so x must be < -1. For negative numbers, x² is positive but x³ is negative, so that means |x²| + |x³| = 36. A quick guess of negative integers shows that x = -3 solves the above, and indeed the original equation as well. Let's see what the other two (almost certainly complex) solutions are. x² - x³ = 36 x³ - x² + 36 = 0 1 -4 12 x+3 | 1 -1 0 36 1 3 -4 0 -4 -12 12 36 12 36 (x+3)(x²-4x+12) = 0 x = -3 | x² - 4x + 12 = 0 x = -(-4)±(√(-4)²-4(1)(12))/2(1) x = 2 ± (1/2)√16-48 x = 2 ± (1/2)√32i = 2 ± (1/2)4√2i x = 2 ± 2√2i x = -3, 2 + 2√2i, 2 - 2√2i
REALLY great! just an other solution... if you are enough brainy to say 36 is 9+27, you are so enough brainy to say -3 is a solution to the problem, so we can factorize by (x-(-3)).... and so this give -x^2, 4x, and -12.... and so......
A couple years ago, abouf a half-dozen Baltimore School District high schools gave diplomas to entire graduating classes which had ZERO students capable of passing standardized 8th grade math tests. So the questions are: 1) Is this REALLY a problem meant for students entering high school, and 2) if it indeed is - which specific schools are we talking about ? And as a bonus question- are these schools in America, or India?
At the time you break 36 to 3^2 and 3^3 , you already know x=-3 is root . And then divide the polynomial by (x+3) , it will be faster.
You explain the beauty in mathematics elegantly.
As a student I didn’t understand that complex numbers are crucial in solving real life problems in physics and engineering.
As a retired engineer mathematics still brings me joy.
x must be < 0. Quick testing yields -3 (-1 and -2 can be excluded at hand and -4 is already to large). Polynomial division by (x+3) yields x² - 4x + 12 which yields two complex solutions 2 +/- 2i*SQRT(2).
Even though I taught math, I don't consider myself a math genius. But -3 just jumped out at me. So even if there is an elaborate solution that nobody ever uses, and few could understand, the -3 is THE answer that applies in the everyday world where we all live.
X needs to be negative so that squaring it becomes positive and cubing becomes negative. That way 'subtracting the negative' becomes 'adding a positive'.
Then just plugged in integers until reaching -3.
x² - x³ = 36
Looking at what any real answers must be, for positive mumbers x² and x³ are both positive, between -1and 1, x² and x³ are also between -1 and 1, and above 1, x³ > x², so x must be < -1. For negative numbers, x² is positive but x³ is negative, so that means |x²| + |x³| = 36. A quick guess of negative integers shows that x = -3 solves the above, and indeed the original equation as well. Let's see what the other two (almost certainly complex) solutions are.
x² - x³ = 36
x³ - x² + 36 = 0
1 -4 12
x+3 | 1 -1 0 36
1 3
-4 0
-4 -12
12 36
12 36
(x+3)(x²-4x+12) = 0
x = -3 | x² - 4x + 12 = 0
x = -(-4)±(√(-4)²-4(1)(12))/2(1)
x = 2 ± (1/2)√16-48
x = 2 ± (1/2)√32i = 2 ± (1/2)4√2i
x = 2 ± 2√2i
x = -3, 2 + 2√2i, 2 - 2√2i
REALLY great! just an other solution... if you are enough brainy to say 36 is 9+27, you are so enough brainy to say -3 is a solution to the problem, so we can factorize by (x-(-3)).... and so this give -x^2, 4x, and -12.... and so......
This is an easier solution by just dividing the known factor with the expressiob!
(x^2)^2 ➖ (3)^2= {x^4 ➖ x^9}= x^5 (x ➖ 5x+5).
Good working and explanation
A couple years ago, abouf a half-dozen Baltimore School District high schools gave diplomas to entire graduating classes which had ZERO students capable of passing standardized 8th grade math tests. So the questions are: 1) Is this REALLY a problem meant for students entering high school, and 2) if it indeed is - which specific schools are we talking about ? And as a bonus question- are these schools in America, or India?
Dirty Sex is really great. 😅
Which high school, I wonder ;-?
I'll try 1, 2 , 3, 4, 5...
You're going to be counting for a long time. The real number solution is -3.
Just plug in values already. I would tell anyone answering as you did, they wasted 10 minutes of their life.
x^2-x^3=36 , x^3 - x^2 +/-x +36 =0 , x^2(x+3)-4x(x+3)+12(x+3)=0 , (x+3)(x^2-4x+12)=0 , x=-3 ,
+1 +3 x^2-4x+12=0 , x=(4+/-V(16-144))/2 , 2+/-iV32 ,
-4 -12 solu, x= -3 , 2+i*V32 , 2-i*V32 ,
+12 +36 =0 , test , (-3)^2-(-3)^3=9-(-27) , 9+27=36 , OK ,
-3 ?😅
Missing 2 other solutions!
HIGH SCHOOL entrance exam? I have friends that are juniors who could not do this shit
wow Great ,but is there any real-life application gor this?
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