Cuberoot Equation

koal · 2003-07-01, 11:50

:? no cube root sign on the web :(

The following equation

CUBEROOT (x + 15) - CUBEROOT (x - 11) = 2

has two (obvious) solutions: x1 = 12, x2 = -16
because CR(27)-CR(1) = 3-1 = 2
and CR(-1)-CR(-27) = -1+3 = 2

But most people (teachers inclusive) have problems to solve it. Who of you can find the tricky way get the results above? 8)

xilman · 2003-07-01, 12:32

I don't know if this is "tricky", but its easy enough. Write the equation as

(x+15)^(1/3) = 2 +(x-11)^1/3)

Write y=(x-11)^(1/3) and cube both sides to get

x+15 = 8 + 12y + 6y^2 + x-11

or 0 = -3 + 2y + y^2

where I've subtracted x+15 from both sides and cleared a common factor of 6.

The quadratic in y is easily seen to factor as (y+3)(y-1) so y = -3 or y=1.

Hence, (x-11)^(1/3) = 1 or (x-11)^(1/3) = -3

Cubing, gives x-11 = 1 or x-11 = -27

Hence, x = 12 or -16 as you said.

What was hard about that?

Paul

hyh1048576 · 2003-07-01, 15:18

Is this tricky enough?

Denote a=(x+15)^(1/3),b=(x-11)^(1/3)
So a-b=2----------------------------(1)
a^3-b^3=26--------------------(2)
(2)/(1): a^2+ab+b^2=13-------(3)
(3)-(1)^2: 3ab=9
So ab=3 and a-b=2
a=-1 or a=3
x+15=-1 or 27
x=-16 or 12

koal · 2003-07-03, 11:58

Correct and both ways tricky, but not the tricky way I ment. I saw students trying to substitute one cube root, saying

y = x-11 <=> x = y+11

then getting

1) CR(Y+26) - CR(Y) = 2

cubing both sides gives

2) Y+26 - 3*CR((Y+26)^2*Y) + 3*CR((Y+26)*Y^2) - Y = 8
or
3) 18 - 3*CR((Y+26)*Y) * (CR(Y+26) - CR(Y)) = 0

So they were lost in the wilderness of cube roots and giving up. But what happens, if one substitutes both cube roots with the arithmetic middle of them?
y = (x+15+x-11)/2
y = x+2 <=> x =y-2 which gives the equation

1) CR(Y+13) - CR(Y-13) = 2

now it's obvious that (see above)

3) 18 - 3*CR((Y+26)*Y) * (CR(Y+26) - CR(Y)) = 0
becomes
3) 18 - 3*CR((Y+13)*(Y-13)) * (CR(Y+13) - CR(Y-13)) = 0

due to 1) we kann substitute CR(Y+13) - CR(Y-13) = 2

4) 18 - 3*CR((Y+13)*(Y-13)*2 = 0
...
5) 3 = CR(Y^2-169)
6) 27 = Y^2-169
7) 196 = Y^2
8) Y1 = 14 <=> x1 = 12 and Y2 = -14 <=> x2 = -16

And I never said, it's hard, I just see, that my "trick" was the longest to write down 8)

2003-07-01, 11:50	#1
koal Nov 2002 Vienna, Austria 51₈ Posts	Cuberoot Equation :? no cube root sign on the web :( The following equation CUBEROOT (x + 15) - CUBEROOT (x - 11) = 2 has two (obvious) solutions: x1 = 12, x2 = -16 because CR(27)-CR(1) = 3-1 = 2 and CR(-1)-CR(-27) = -1+3 = 2 But most people (teachers inclusive) have problems to solve it. Who of you can find the tricky way get the results above? 8)

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2003-07-01, 12:32	#2
xilman Bamboozled! "𒉺𒌌𒇷𒆷𒀭" May 2003 Down not across 2DFB₁₆ Posts	I don't know if this is "tricky", but its easy enough. Write the equation as (x+15)^(1/3) = 2 +(x-11)^1/3) Write y=(x-11)^(1/3) and cube both sides to get x+15 = 8 + 12y + 6y^2 + x-11 or 0 = -3 + 2y + y^2 where I've subtracted x+15 from both sides and cleared a common factor of 6. The quadratic in y is easily seen to factor as (y+3)(y-1) so y = -3 or y=1. Hence, (x-11)^(1/3) = 1 or (x-11)^(1/3) = -3 Cubing, gives x-11 = 1 or x-11 = -27 Hence, x = 12 or -16 as you said. What was hard about that? Paul

2003-07-01, 15:18	#3
hyh1048576 Jun 2003 2⁶ Posts	Is this tricky enough? Denote a=(x+15)^(1/3),b=(x-11)^(1/3) So a-b=2----------------------------(1) a^3-b^3=26--------------------(2) (2)/(1): a^2+ab+b^2=13-------(3) (3)-(1)^2: 3ab=9 So ab=3 and a-b=2 a=-1 or a=3 x+15=-1 or 27 x=-16 or 12

2003-07-03, 11:58	#4
koal Nov 2002 Vienna, Austria 41 Posts	Correct and both ways tricky, but not the tricky way I ment. I saw students trying to substitute one cube root, saying y = x-11 <=> x = y+11 then getting 1) CR(Y+26) - CR(Y) = 2 cubing both sides gives 2) Y+26 - 3CR((Y+26)^2Y) + 3CR((Y+26)Y^2) - Y = 8 or 3) 18 - 3CR((Y+26)Y) * (CR(Y+26) - CR(Y)) = 0 So they were lost in the wilderness of cube roots and giving up. But what happens, if one substitutes both cube roots with the arithmetic middle of them? y = (x+15+x-11)/2 y = x+2 <=> x =y-2 which gives the equation 1) CR(Y+13) - CR(Y-13) = 2 now it's obvious that (see above) 3) 18 - 3CR((Y+26)Y) * (CR(Y+26) - CR(Y)) = 0 becomes 3) 18 - 3CR((Y+13)(Y-13)) * (CR(Y+13) - CR(Y-13)) = 0 due to 1) we kann substitute CR(Y+13) - CR(Y-13) = 2 4) 18 - 3CR((Y+13)(Y-13)*2 = 0 ... 5) 3 = CR(Y^2-169) 6) 27 = Y^2-169 7) 196 = Y^2 8) Y1 = 14 <=> x1 = 12 and Y2 = -14 <=> x2 = -16 And I never said, it's hard, I just see, that my "trick" was the longest to write down 8)