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drek231 [11]
3 years ago
15

Solve the equation x^13-2x^12-x^11+2x^10=0 in the real number system

Mathematics
1 answer:
ASHA 777 [7]3 years ago
3 0

Answer:

x ={0; - 1 ;  2 ;  1}

Step-by-step explanation:

x¹³-2x¹²-x¹¹+2x¹⁰=0

x¹⁰(x³-2x²-x+2)=0

x¹⁰(x+1)(x-2)(x-1)=0

x¹⁰=0    ∨   x+1=0    ∨    x-2=0   ∨   x-1=0

x=0             x= - 1             x= 2          x= 1

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WHY IS THIS GETTING DELETED?? I NEED HELP
aliina [53]

If this exact question is repeatedly deleted, it's probably because of the ambiguity of the given equation. I see two likely interpretations, for instance:

\dfrac{(5\times5)^k}{5^{-8}} = 5^3

or

\dfrac{5\times 5^k}{5^{-8}} = 5^3

If the first one is what you intended, then

\dfrac{(5\times5)^k}{5^{-8}} = \dfrac{(5^2)^k}{5^{-8}} = \dfrac{5^{2k}}{5^{-8}} = 5^{2k-(-8)} = 5^{2k+8} = 5^3

and it follows that

2<em>k</em> + 8 = 3   ==>   2<em>k</em> = -5   ==>   <em>k</em> = -5/2

If you meant the second one, then

\dfrac{5\times 5^k}{5^{-8}} = \dfrac{5^1\times5^k}{5^{-8}} = \dfrac{5^{k+1}}{5^{-8}} = 5^{k+1-(-8)} = 5^{k+9} = 5^3

which would give

<em>k</em> + 9 = 3   ==>   <em>k</em> = -6

And for all I know, you might have meant some other alternative... When you can, you should include a picture of your problem.

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2 years ago
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