- #1
anemone
Gold Member
MHB
POTW Director
- 3,883
- 115
Which is larger, the real root of x + x2 + ... + x8 = 8 - 10x9, or the real root of x + x2 + ... + x10 = 8 - 10x11?
HallsofIvy said:The second one. The first has root about 0.882, the second, bout 0.884. I got that by graphing both and then "zooming" in on the zeros.
Jester said:Here's my solution
It's not hard to show that the derivative of each is positive so both are increasing function meaning there's only one root. Let
$f(x) = 1 + x + x^2+ \cdots +x^8 + 10x^9 - 8$
and
$g(x) = 1 + x + x^2+ \cdots + x^{10} + 10x^{11} - 8$
We also have $f(0.8) <0, \; f(0.9) >0, \;g(0.8) <0, \;g(0.9) >0$ meaning that both roots lie between $0.8$ and $0.9$. Now consider the difference $h(x) = f(x)-g(x) =9x^8-x^{10}-10x^{11}= -(x+1)(10x-9)x^9 $. On the interval $(0.8,0.9)$ $h(x)> 0$ meaning that $ f(x) > g(x)$ giving that the root of $g(x)$ is larger than the root of $f(x)$ as HallsofIvy pointed out.
anemone said:Which is larger, the real root of x + x2 + ... + x8 = 8 - 10x9, or the real root of x + x2 + ... + x10 = 8 - 10x11?
$1-x^9=(1-x)(x+x^2+\cdots+x^8)$ $1-x^9=(1-x)(8-10x^9)$ $1-x^9=8-10x^9-8x+10x^{10}$ $10x^{10}-9x^9-8x+7=0$ | $1-x^{11}=(1-x)(x+x^2+\cdots+x^{10})$ $1-x^{11}=(1-x)(8-10x^{11})$ $1-x^{11}=8-10x^{11}-8x+10x^{12}$ $10x^{12}-9x^{11}-8x+7=0$ |
So I let $f(x)=10x^{10}-9x^9-8x+7$ | So I let $g(x)=10x^{12}-9x^{11}-8x+7$ |
Descartes's Rule says $f(x)$ has two positive real roots and obviously $x=1$ is one of the root of $f(x)$ and the other root lies between $(0, 1)$. | Descartes's Rule says $g(x)$ has two positive real roots and obviously $x=1$ is one of the root of $g(x)$ and the other root lies between $(0, 1)$. |
A cube has a larger root than a square. This is because a cube has three dimensions, while a square only has two. Therefore, the cube's root is the number that, when multiplied by itself three times, gives the volume of the cube.
To determine which has a larger root, you can simplify the expressions and compare the numbers. If the simplified expressions have the same base, you can compare their exponents. If the numbers have different bases, you can use a calculator to find their square roots and compare them.
Yes, you can approximate the roots of irrational numbers using a calculator or by using the long division method. However, it is important to keep in mind that the roots of irrational numbers are non-terminating and non-repeating decimals, so they cannot be compared precisely.
Yes, the roots of negative numbers can be compared. However, it is important to note that when taking the square root of a negative number, the result is a complex number. Therefore, the comparison would be between two complex numbers.
Yes, there are some special cases where the larger root may not be obvious. For example, when comparing two perfect squares, they will have the same root. Similarly, when comparing two perfect cubes, they will have the same root. In these cases, the root is not larger or smaller, but equal.