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3 years ago

A polynomial function P(x) with rational coefficients has the roots sqrt 2 and - sqrt 13. Find two additional roots.

Mathematics

1 answer:

bixtya [17]3 years ago

4 0

Answer:

$-\sqrt{2}$ and $\sqrt{13}$

Step-by-step explanation:

Given a polynomial with rational coefficients, its irrational and complex roots would occur in conjugate pairs.

For example, if the roots of a polynomial are 2 + $\sqrt{3}$ , 3 and 4 - 3i, then 2 - $\sqrt{3}$ , and 4 + 3i are also roots of the polynomial.

Therefore, since the given polynomial has rational coefficients and some of its roots are , $\sqrt{2}$ and - $\sqrt{13}$ , then - $\sqrt{2}$ and $\sqrt{13}$ are also roots of the polynomial.

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Camille is attending a fundraiser. She pays for her admission and buys raffle tickets for $ 5 $5 each. If she buys 1 0 10 raffle

tatyana61 [14]

The question is a little confusing, but I think the answer goes like this:
135 = admission + 5x, and if she bought 10 tickets:
135 = admission + 50, so admission must have cost her $85
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3 years ago

How do I do this ? I have 120 degrees but I need to solve for x

Darina [25.2K]

Since the legs are the same length, we know that the remaining two angles must be the same.

120+2x=180
2x=60
x=30

Final answer: x=30 degrees

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3 years ago

Rob is saving to buy a new MP3 player. For every $13 he earns babysitting, he saves $8. On Saturday, Rob earned $65 babysitting.

Vedmedyk [2.9K]

Answer:

Step-by-step explanation:

So you first start off with 13 and 8 you subtract it and get five so he uses 5 every time he gets paid and you subtract 5 from 65 and get 60

--Hope this helps

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3 years ago

Solve these recurrence relations together with the initial conditions given. a) an= an−1+6an−2 for n ≥ 2, a0= 3, a1= 6 b) an= 7a

8_murik_8 [283]

Answer:

a) 3/5·((-2)^n + 4·3^n)
b) 3·2^n - 5^n
c) 3·2^n + 4^n
d) 4 - 3 n
e) 2 + 3·(-1)^n
f) (-3)^n·(3 - 2n)
g) ((-2 - √19)^n·(-6 + √19) + (-2 + √19)^n·(6 + √19))/√19

Step-by-step explanation:

These homogeneous recurrence relations of degree 2 have one of two solutions. Problems a, b, c, e, g have one solution; problems d and f have a slightly different solution. The solution method is similar, up to a point.

If there is a solution of the form $a[n]=r^n$ , then it will satisfy ...

$r^n=c_1\cdot r^{n-1}+c_2\cdot r^{n-2}$

Rearranging and dividing by $r^{n-2}$ , we get the quadratic ...

$r^2-c_1r-c_2=0$

The quadratic formula tells us values of r that satisfy this are ...

$r=\dfrac{c_1\pm\sqrt{c_1^2+4c_2}}{2}$

We can call these values of r by the names r₁ and r₂.

Then, for some coefficients p and q, the solution to the recurrence relation is ...

$a[n]=pr_1^n+qr_2^n$

We can find p and q by solving the initial condition equations:

$\left[\begin{array}{cc}1&1\\r_1&r_2\end{array}\right] \left[\begin{array}{c}p\\q\end{array}\right] =\left[\begin{array}{c}a[0]\\a[1]\end{array}\right]$

These have the solution ...

$p=\dfrac{a[0]r_2-a[1]}{r_2-r_1}\\\\q=\dfrac{a[1]-a[0]r_1}{r_2-r_1}$

_____

Using these formulas on the first recurrence relation, we get ...

$c_1=1,\ c_2=6,\ a[0]=3,\ a[1]=6\\\\r_1=\dfrac{1+\sqrt{1^2+4\cdot 6}}{2}=3,\ r_2=\dfrac{1-\sqrt{1^2+4\cdot 6}}{2}=-2\\\\p=\dfrac{3(-2)-6}{-5}=\dfrac{12}{5},\ q=\dfrac{6-3(3)}{-5}=\dfrac{3}{5}\\\\a[n]=\dfrac{3}{5}(-2)^n+\dfrac{12}{5}3^n$

The rest of (b), (c), (e), (g) are solved in exactly the same way. A spreadsheet or graphing calculator can ease the process of finding the roots and coefficients for the given recurrence constants. (It's a matter of plugging in the numbers and doing the arithmetic.)

_____

For problems (d) and (f), the quadratic has one root with multiplicity 2. So, the formulas for p and q don't work and we must do something different. The generic solution in this case is ...

$a[n]=(p+qn)r^n$

The initial condition equations are now ...

$\left[\begin{array}{cc}1&0\\r&r\end{array}\right] \left[\begin{array}{c}p\\q\end{array}\right] =\left[\begin{array}{c}a[0]\\a[1]\end{array}\right]$

and the solutions for p and q are ...

$p=a[0]\\\\q=\dfrac{a[1]-a[0]r}{r}$

Using these formulas on problem (d), we get ...

$c_1=2,\ c_2=-1,\ a[0]=4,\ a[1]=1\\\\r=\dfrac{2+\sqrt{2^2+4(-1)}}{2}=1\\\\p=4,\ q=\dfrac{1-4(1)}{1}=-3\\\\a[n]=4-3n$

And for problem (f), we get ...

$c_1=-6,\ c_2=-9,\ a[0]=3,\ a[1]=-3\\\\r=\dfrac{-6+\sqrt{6^2+4(-9)}}{2}=-3\\\\p=3,\ q=\dfrac{-3-3(-3)}{-3}=-2\\\\a[n]=(3-2n)(-3)^n$

_____

<em>Comment on problem g</em>

Yes, the bases of the exponential terms are conjugate irrational numbers. When the terms are evaluated, they do resolve to rational numbers.

6 0

2 years ago

Where are the asymptotes for the following function located? f(x)= 7/ x^2-2x-24

Zarrin [17]

Factor the demoniator:-

x^2 -2x - 24 = (x - 6)(x + 4)

the asymptotes occurs when denominator = 0

so here they are the vertical lines x = 6 and x = -4

7 0

3 years ago

Read 2 more answers