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anemone
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Find all irrational numbers $k$ such that $k^3-17k$ and $k^2+4k$ are both rational numbers.
mathbalarka said:$k^2 + 4k = x$ be rational, following the question. Multiplying by $k$, we get $k^3 + 4k^2 = xk$. Substituting $k^2 = x - 4k$ in, one gets $k^3 + 4(x - 4k) = xk$. Thus, $k^3 - 16k = xk - 4x$ which implies $k^3 - 17k = xk - 4x - k = (x - 1)k - 4x$. But this thingy $k^3 - 17k$ is rational as per the question, which implies $(x - 1)k - 4x$ is rational. But $k$ is irrational and $x - 1$ and $4x$ are both rational ($x$ being rational). That'd imply $(x - 1)k - 4x$ is irrational, a contradiction if and only if $x - 1$ is nonzero. Otherwise, $x = 1$ and hence the desired irrationals are the roots of $k^2 + 4k = 1$.
Opalg said:[sp]Let $k^2+4k = r$, rational. The quadratic equation $k^2+4k -r=0$ has solutions $k = -2 \pm\sqrt{4+r}$. Then $k^2 = 8+r \mp\sqrt{4+r}$ and $$k^3 - 17k = k(k^2-17) = (-2 \pm\sqrt{4+r})(-9+r \mp\sqrt{4+r}) = -2 -6r \pm(r-1)\sqrt{4+r}.$$ For that to be rational, we must have $r=1$ (because $\sqrt{4+r}$ must be irrational). That gives the solutions $k = -2\pm\sqrt5.$[/sp]
kaliprasad said:$k^2+4k$ is rational so $(k^2+ 4k + 4)$ or $(k+2)^2$
so $k = -2 \pm \sqrt{q}$ where q is rational but not square of rational
take $k = \sqrt{q}-2$ and
so we get
$k^3-17k$
= $(\sqrt{q} - 2)^3- 17(\sqrt{q}- 2)$
now take the irrational part to be zero to get
$q\sqrt{q} + 12\sqrt{q} - 17\sqrt{q}=0$ or q = 5
so one solution $\sqrt(5) - 2$
taking -ve q we get $-2-\sqrt{5}$
the 2 solutions are $\sqrt(5) - 2$ and $-2-\sqrt{5}$
Even though your conclusion is wrong, if you want to re-check your approach, just know that you're always welcome to re-submit of your most satisfied solution!
mathbalarka said:Ahem. I don't think my conclusion is wrong. The roots of $k^2 + 4k - 1 = 0$ are precisely $-2 \pm \sqrt{5}$, which satisfied your original conditions.
An irrational number is a number that cannot be expressed as a ratio of two integers. These numbers cannot be written as terminating or repeating decimals and their decimal representation continues infinitely without any pattern.
To find all irrational numbers k, you can use various methods such as the Pythagorean theorem, the square root method, or the continued fraction method. These methods involve finding the square root of a non-perfect square number or by using a formula to generate a sequence of numbers that approach an irrational number.
No, it is impossible to find all irrational numbers as there are infinitely many of them. Moreover, new irrational numbers are being discovered every day, making it impossible to list or find all of them.
No, there are no patterns in the decimal representation of irrational numbers. Unlike rational numbers, which have repeating or terminating decimals, irrational numbers have an infinite and non-repeating decimal representation.
Some real-life examples of irrational numbers include pi (π), which is the ratio of a circle's circumference to its diameter, and the square root of 2, which represents the diagonal of a square with sides of length 1 unit. These numbers cannot be expressed as a ratio of two integers and are used in various mathematical calculations and formulas.