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3 years ago

2x^2+3x+5=0 what are all the complex solutions?please i need help!!!

Mathematics

2 answers:

blagie [28]3 years ago

7 0

2x^2+3x+5=0

Step 1: Use quadratic formula with a=2, b=3, c=5 ^^ than do ^^

There is no real solutions I'm pretty sure!

Hope this helped!!

vesna_86 [32]3 years ago

4 0

You'll want to use the quadratic formula:

-b (+/-) sqrt(b^2 - 4ac), all divided by 2a.

Under the square root you'll get:

-11

remember that the square root of -1 is i.

sqrt(-11) can be factored to sqrt(11*-1) and then sqrt(-1) * sqrt(11)

which becomes i*sqrt(11)

so your complex solution is:

-3 (+/-) (i*sqrt(11)), all over 4

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2 years ago

F(x) = 3x + 6 please help quickly!!!!!!!

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Given the function, f(x) = 3x + 6, we can solve for f(a), f(a + h) and $\frac{(f(a + h) - f(a)) }{h}$ by substituting their values into f(x) = 3x + 6. We will have the following:

$\mathbf{f(a) = 3a + 6}\\\\\mathbf{f(a + h) = 3a + 3h + 6}\\\\\mathbf{\frac{(f(a + h) - f(a)) }{h} = 6}$

Given:

f(x) = 3x + 6

We are told to find:

f(a)
f(a + h), and
$\frac{(f(a + h) - f(a)) }{h}$

1. Find f(a):

Substitute x = a into f(x) = 3x + 6

f(a) = 3(a) + 6

f(a) = 3a + 6

2. Find f(a + h):

Substitute x = a + h into f(x) = 3x + 6

f(a + h) = 3(a + h) + 6

f(a + h) = 3a + 3h + 6

3. Find $\frac{(f(a + h) - f(a)) }{h}$ :

Plug in the values of f(a + h) and f(a) into $\frac{(f(a + h) - f(a)) }{h}$

Thus:

$\frac{((3a + 3h + 6) - (3a + 6)) }{h}\\\\\frac{(3a + 3h + 6 - 3a - 6) }{h}\\\\$

Add like terms

$\frac{(3a - 3a + 3h + 6 - 6) }{h}\\\\= \frac{3h }{h}\\\\\mathbf{= 3}$

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$\mathbf{f(a) = 3a + 6}\\\\\mathbf{f(a + h) = 3a + 3h + 6}\\\\\mathbf{\frac{(f(a + h) - f(a)) }{h} = 6}$

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