Japanese exam math question

[@KorlinAng-bs7rh]

let y=x^2-2 and y^2=x+2, added the 2 equations, we get (x-y)(x+y+1)=0. so 2 cases will produce the 4 solutions.

[@mcwulf25]

Well done. I was going to go down the system of equations route but you saved me the trouble!

[@johnschultz4345]

This seems like another math trivia question where you either just know the magic substitution that makes everything work out or you are SOL. For a slightly less contrived approach, I leveraged the obvious root of x = 2 of the original equation and used polynomial long division on the quartic equation: x^2 - 2 = sqrt(x + 2) x^4 - 4x^2 - x + 2 = 0 (x - 2)(x^3 + 2x^2 - 1) = 0 If you look at the resulting cubic expression, then you can find another obvious root of x = -1 and use polynomial long division again: (x - 2)(x + 1)(x^2 + x - 1) = 0 Then you can use the quadratic formula to find the 3rd and 4th roots of x = [-1 +/- sqrt(5)] / 2. Then you can check your roots with the original equation. This approach only "just so happens to work" because two of the roots were integers, easily guessed + checked, and rather obvious.

[@mcwulf25]

100%. This is often the way we factor polynomials anyway. In exam questions, there is often a solution that avoids pages of algebra!

[@actualRocketScientist]

I think you can skip a bunch of steps if you first square then pull out an x squared and then apply the difference of two squares and then combine appropriate quantities

[@muontran8589]

You let t=2; so you did too long,crazy!!!!

[@Penndennis]

A quadratic in terms of 2 - functions as both a variable and a directly calculable number. Excellent stuff!

[@dhpbear2]

4:56 - Why did you change "-1" to "+1"?

[@prollysine]

we get , x^4+/-x^3-4x^2-x+2=0 , (x-2)(x^3+2x^2-1)=0 , x^3+2x^2+/-x-1=0 , (x+1)(x^2+x-1)=0 , x^2+x-1=0 , x=(-1+/-V5)/2 , 1 -2 -1 2 1 1 result , x= 2 , -1 , (-1+V5)/2 , (-1-V5)/2 , 2 -4 1 1 test --> x= -1 & (-1+V5)/2 not a solu , -1 -1 solu , x= 2 , (-1-V5)/2 , OK ,

[@TheMathManProfundities]

x²-2 = √(x+2) ⇒ x⁴ - 4x² + 4 = x + 2 ⇒ x⁴ - 4x² - x + 2 = 0. Try simple integers, observe x=-1 or 2. (x+1)(x-2) = x² - x - 2 so (x²-x-2)(x²+x-1) = 0. For other potential solutions, x² + x - 1 = 0 ⇒ x = {-1 ± √(1² - 4×1×-1)}/2 = (-1±√5)/2. All potential solutions are real so x²-2 ≥ 0 and x+2 ≥ 0 so we can reject -1 and (-1+√5)/2. Check x= 2 and (-1-√5)/2 against original equation: x=2: x²-2 = (2)² - 2 = 2 = √{(2)+2} = √(x+2) Valid. x=(-1-√5)/2: x²-2 = {(-1-√5)/2}² - 2 = (1+5+2√5)/4 - 2 = (-1 +√5)/2 ≥ 0 Valid. So x = 2 or (-1-√5)/2.

[@log-i8g]

x=[- φ, round φ]

[@omega3li493]

Square both side. Done... Dimple grade-9 Maths... 哈哈！

[@ankitpatel-lm2in]

-1 is possible. No need to reject it. In the beginning the notion is adopted that x+2 should be greater than 0. Yes. But that doesn't mean the square roof of x+2 cannot be a negative number. The square roof of 1 can be -1 or 1. If you replace x with -1, it will be true for both sides.

[@RealQinnMalloryu4]

(x^2)^2 ➖ (2)^2={x^4 ➖ 4}={x^0+x^0 ➖ x^0+x^0 ➖ x^0+x^0 ➖ x^0+x^0 ➖ }={x^1+x^1+x^1+x^1}=x^4 x^2^2 (x ➖ 2x+2) {x+x ➖}+{2+2 ➖}={x^2+4}=4x^2 2^2x^2 1^1x^2 1x^2 (x ➖ 2x+1).

[@charl10439]

X=2

[@Cernl.1]

X=6.686363😊

[@charl10439]

2

[@venkybabu4842]

A square root has 2 solutions. A cube root 4.

[@mcwulf25]

But a square root sign means just the positive root, and a cubic has 3.

[@sarahpierris3004]

lower mathematics

[@AyushKalmodiya-j3g]

X=2 😂

[@NounPhechnai]

Second

[@Moulisantu]

X=2