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a_sh-v [17]
3 years ago
10

Can you please find the zeros for F(x)=2x^4+17x^2+8?

Mathematics
1 answer:
harina [27]3 years ago
4 0
2x⁴+17x²+8

Let X = x²     →→(x = + or -√X )

The quadratic becomes : 2.X² + 17.X + 8. Now solve for X:

X' = [-b+√(b²-4ac)]/2a             and      X" = [-b-√(b²-4ac)]/2a   
Plug I the values of a, b and c and you will find;

X' = -1/2   and X" = - 8
Since X = √x → →X' = √( -1/2) [imaginary roots]
and X" = √-8 →→ X" = 2√-2  [imaginary roots]
Then:

1st ROOT: X' = - √1/2.i  
2nd ROOT: X" = + √1/2.i
3rd ROOT: X" = + 2√2.i
4th ROOT: X" = - 2√2.i


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Answer:

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Step-by-step explanation:

We have the tables:

<u>Table 1:</u>

x: 1  2  3  4

y: 2  4  6  8

<u>Table 2:</u>

x: 1  2  3  4

y: 2  4  8  16

<u>Table 3:</u>

x: 1  2  3  4

y: 2  4  7  11

<u>Table 4:</u>

x: 1  2  3  4

y: 2  4  6  10

An exponential growth data set will show a common ratio between y values. Let's look at each of the ratios from each table.

<u>Table 1:</u>

8/6 = 4/3

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Already, we can see that 4/3 ≠ 3/2, which means that this doesn't have a common ratio. So Table 1 is wrong.

<u>Table 2:</u>

16/8 = 2

8/4 = 2

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The common ratio here is 2, so we know this is correct.

<u>Table 3:</u>

11/7 = 1.57

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Again, we can see that 1/57 ≠ 1.75, so this is wrong.

<u>Table 4:</u>

10/6 = 1.67

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Again, there is no common ratio here, so this is wrong.

The answer is thus Table 2.

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