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nata0808 [166]
3 years ago
5

Slove x^2-8x=20 by completing the square. which is the solution set of equation?​

Mathematics
1 answer:
sveticcg [70]3 years ago
3 0

x^2-8x=20

Move +20 to the other side. Sign changes from +20 to -20

x^2-8x-20=20-20

x^2-8x-20=0

Think about two numbers that equals to -20 and -8.

(x-10)(x+2)=0

x-10=0 ( move -10 to the other side). Sign changes from -10 to 10

x-10+10=0+10

x=10

x+2=0 ( move +2 to the other side). Sign changes from +2 to -2

x+2-2=0-2

x=-2

Answer:

x=10, x=-2

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Determine formula of the nth term 2, 6, 12 20 30,42​
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Check the forward differences of the sequence.

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b_n = a_{n+1} - a_n

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Let \{c_n\} be the sequence of differences of \{b_n\},

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2 = b_{n+1} - b_n \implies b_{n+1} = b_n + 2

and we can solve for b_n in terms of b_1=4:

b_{n+1} = b_n + 2

b_{n+1} = (b_{n-1}+2) + 2 = b_{n-1} + 2\times2

b_{n+1} = (b_{n-2}+2) + 2\times2 = b_{n-2} + 3\times2

and so on down to

b_{n+1} = b_1 + 2n \implies b_{n+1} = 2n + 4 \implies b_n = 2(n-1)+4 = 2(n + 1)

We solve for a_n in the same way.

2(n+1) = a_{n+1} - a_n \implies a_{n+1} = a_n + 2(n + 1)

Then

a_{n+1} = (a_{n-1} + 2n) + 2(n+1) \\ ~~~~~~~= a_{n-1} + 2 ((n+1) + n)

a_{n+1} = (a_{n-2} + 2(n-1)) + 2((n+1)+n) \\ ~~~~~~~ = a_{n-2} + 2 ((n+1) + n + (n-1))

a_{n+1} = (a_{n-3} + 2(n-2)) + 2((n+1)+n+(n-1)) \\ ~~~~~~~= a_{n-3} + 2 ((n+1) + n + (n-1) + (n-2))

and so on down to

a_{n+1} = a_1 + 2 \displaystyle \sum_{k=2}^{n+1} k = 2 + 2 \times \frac{n(n+3)}2

\implies a_{n+1} = n^2 + 3n + 2 \implies \boxed{a_n = n^2 + n}

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2 years ago
Find the sum or difference in -8+13=
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Answer:

5

Step-by-step explanation:

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