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alexira [117]
3 years ago
8

A polynomial equation with rational coefficients has the roots 7 + *sqrt* 3 , 2 - *sqrt* 6. Find two additional roots.

Mathematics
2 answers:
topjm [15]3 years ago
7 0
If the coefficients of a polynomial equation are rational, then the irrational roots have to come in conjugate pairs.
The conjugate of 7 +√ 3 is 7 - √ 3 and the conjugate of 2 - √ 6 is 2 + √ 6.
Answer:
A ) 7 - √ 3 ,  2 + √ 6
Ganezh [65]3 years ago
5 0

Answer:

option: A

Step-by-step explanation:

" if for a polynomial with rational coefficients has irrational roots then these irrational roots will always appear in pair " .

we are given that 7+\sqrt{3} and 2-\sqrt{6} are two roots of a polynomial equation with rational coefficients, then it's complex conjugate is also a root of this polynomial equation i.e. ' 7-\sqrt{3} ' and '  2+\sqrt{6} ' are also roots of this polynomial equation.

Hence, option A is correct.


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Answer:

first answer is correct

Step-by-step explanation:

Hello, we want to write 2 + 4 + 6 + ... + 20

for n = 1, 2n = 2*1 =2

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\displaystyle \sum_{n=1}^{10} {2n} = 2*1 + 2*2 +2*3+2*4+...+2*10=2+4+6+8+...+20

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3 years ago
The sum of two numbers is 5000. if 20% of one number is equal to 30% of other no. find the number
Alex Ar [27]
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7 0
4 years ago
A candle shop sells scented candles for $16 each and unscented candles for $10 each.The shop sells 28 candles today and makes $4
Greeley [361]
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3 years ago
Read 2 more answers
Write the equation of the line that passes through the points
Marrrta [24]

Answer:

y+2-x=0

Step-by-step explanation:

Here,

(-3,-5)=(x1,y1)

(3,1)=(x2,y2)

Slope(m)=y2-y1/x2-x1

=1-(-5)/3-(-3)

=1+5/3+3

=6/6

=1

Mid point=(x1+x2/2 , y1+y2/2)

=(-3+3/2 , -5+1/2)

=(0/2 , -4/2)

=(0 , -2)

Now, (0,-2)=(x1,y1)

Equation:

y-y1=m(x-x1)

y+2=1(x-0)

y+2=1x

y+2-x=0 is the required equation.

3 0
3 years ago
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