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Is the given quantity an integer?

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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,683
Your calculator tells you that the quantity $y=(2\cdot\sqrt[3]{2}+1-\sqrt{12\cdot\sqrt[3]{2}-15})^3$ is approximately an integer. Is $y$ exactly an integer? Justify your answer.
 

kaliprasad

Well-known member
Mar 31, 2013
1,309
Your calculator tells you that the quantity $y=(2\cdot\sqrt[3]{2}+1-\sqrt{12\cdot\sqrt[3]{2}-15})^3$ is approximately an integer. Is $y$ exactly an integer? Justify your answer.
yes it is 32


as

Because under the square root we have $\sqrt[3]{2}$
It suggests that the square root is
a + b $\sqrt[3]{2}$ + c $\sqrt[3]{4}$
And taking square we get
$a^2+ b^2\sqrt[3]{4} +2c^2\sqrt[3]{4} +2ab\sqrt[3]{2} + 4bc + 2ac\sqrt[3]{4} = 12\sqrt[3]{2}−15$
Equating the parts that is rational , $\sqrt[3]{2}$ and $\sqrt[3]{4}$
We get
$a^2 + 4bc = - 15$
$b^2 + 2ac = 0$
$c^2 + ab = 6$
Solving these ( I do not know how to solve but guess work a = 1, b= 2, c =- 2
so square root = 1 + 2 $\sqrt[3]{2}$ - 2 $\sqrt[3]{4}$

so given expression is 2 $\sqrt[3]{4}$ which is cubed to give 32
 

Albert

Well-known member
Jan 25, 2013
1,225
Your calculator tells you that the quantity $y=(2\cdot\sqrt[3]{2}+1-\sqrt{12\cdot\sqrt[3]{2}-15})^3$ is approximately an integer. Is $y$ exactly an integer? Justify your answer.

let $\sqrt[3]{2}=x$
we have :$x^3=2---(1)$
$y= (2x+1-\sqrt {12x-15})^3$
if y is an integer then we may suggest :
$\sqrt {12x-5}=2x+1-ax^2---(2)$
(here a is an integer)
square both sides of (2) and use of (1) we get a=2
$\therefore y=(2x^2)^3=32$
 
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anemone

MHB POTW Director
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Feb 14, 2012
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Thank you all for participating and sorry for the late reply. :eek:

Suggested solution by other:

According to the calculator, $y$ is approximately equal to 32, and so we would like to decide whether or not the quantity $x=2\cdot\sqrt[3]{2}+1-\sqrt{12\cdot\sqrt[3]{2}-15}$ is exactly equal to $\sqrt[3]{32}$.

For notational simplicity, let $a=\sqrt[3]{2}$ so that $x=2a+1-\sqrt{12a-15}$.

Also, since $32=2^5$, we see that $\sqrt[3]{32}=2a^2$ and hence we need to determine whether or not the equation $\sqrt{12a-15}=2a+1-2a^2$ is valid.

Using calculator, we see that the right side of this equation exceeds 0.3, and so it is definitely positive. We can thus check the equation by showing that the squares of both sides are equal. If we square the right side and combine like terms, we get $4a^4-8a^3+4a+1$. But since $a^3=2$, we see that $a^4=2a$ and thus $4a^4-8a^3+4a+1=12a-15$. Consequently, our equation is true, and $y$ is indeed exactly equal to 32.
 
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