Can You Pass High School Entrance Exam?

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[@wu2carlos]

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.

[@Segalmed]

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).

[@sarahpierris3004]

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......

[@Teraverse420]

This is an easier solution by just dividing the known factor with the expressiob!

[@robertloveless4938]

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.

[@robertloveless4938]

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?

[@dieterkohler2206]

Another one of those "solutions" where you know the answer and work backwards to figure out a solution trick.... That's not math, thats "Guess the solution first". The general solution is you set (x - a) + (x^2 + bc + c) = 0 and then compare the coefficients of x^3, x^2, x^1, and x^0 which gives you a,b,c and the solutions to the problem.

[@dieterkohler2206]

small correction, not + but * : (x - a) * (x^2 + bc + c) = 0

[@quigonkenny]

Using logic to narrow the real solution down to -3 first seems a lot faster to me, and still uses principles of math. x² - x³ < 0 for all x > 1, and for all -1 ≥ x ≥ 1, -1 ≥ x², x³ ≥ 1, so x < -1. |x³| < 36, so the only possible integer solutions are -3 and -2, and it's simple to both confirm the former and eliminate the latter. From there pure math gets you the complex solutions rather simply.

[@bonesdmz]

@@quigonkennyput 35 in there, and all this doesn’t work. Solution has to work for each and every number out there, not only for one with “miraculously” found integer root

[@magnusmagato]

Can you explain your solution in more detail please? What's the name of this approach?

[@quigonkenny]

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

[@Watcher40K9]

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.

[@RealQinnMalloryu4]

(x^2)^2 ➖ (3)^2= {x^4 ➖ x^9}= x^5 (x ➖ 5x+5).

[@dprcontracting6299]

Good working and explanation

[@dakershaffer_blitz]

Which high school, I wonder ;-?

[@davidwhiteley3879]

Just plug in values already. I would tell anyone answering as you did, they wasted 10 minutes of their life.

[@mr.sheldor794]

HIGH SCHOOL entrance exam? I have friends that are juniors who could not do this shit

[@moonlight_3141]

I'll try 1, 2 , 3, 4, 5...

[@ericwelsh4853]

You're going to be counting for a long time. The real number solution is -3.

[@NothingMaster]

Dirty Sex is really great. 😅

[@prollysine]

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 ,

[@quickwashtheraccoon7601]

-3 ?😅

[@ruschein]

Missing 2 other solutions!

[@anindabanik208]

wow Great ,but is there any real-life application gor this?

[@likhon007-c3n]

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