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2 years ago

Which are the solutions of x2 = 19x + 1?

Mathematics

2 answers:

Finger [1]2 years ago

6 0

Answer:

$(\frac{19-\sqrt{365}} {2},\frac{19+\sqrt{365}} {2})$

Step-by-step explanation:

we have

$x^2=19x+1$

we know that

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

$x=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$-x^{2}-19x-1=0$

$a=1\\b=-19\\c=-1$

substitute in the formula

$x=\frac{-(-19)\pm\sqrt{-19^{2}-4(1)(-1)}} {2(1)}$

$x=\frac{19\pm\sqrt{365}} {2}$

$x=\frac{19+\sqrt{365}} {2}$

$x=\frac{19-\sqrt{365}} {2}$

$(\frac{19-\sqrt{365}} {2},\frac{19+\sqrt{365}} {2})$

therefore

StartFraction 19 minus StartRoot 365 EndRoot Over 2 EndFraction comma StartFraction 19 + StartRoot 365 EndRoot Over 2 EndFraction

masya89 [10]2 years ago

3 0

Answer:

The answer is B

Step-by-step explanation:

I took the test

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Answer:

Check attachment for complete question

Step-by-step explanation:

Given that,

y=Coskt

We are looking for value of k, that satisfies 4y''=-25y

Let find y' and y''

y=Coskt

y'=-kSinkt

y''=-k²Coskt

Then, applying this 4y'"=-25y

4(-k²Coskt)=-25Coskt

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Divide through by Coskt and we assume Coskt is not equal to zero

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Since k= ± 5/2

y=A•Sin(±5/2t)+ B •Cos(±5/2t)

y'=±5/2ACos(±5/2t)-±5/2BSin(±5/2t)

y''=-25/4ASin(±5/2t)-25/4BCos(±5/2t

Then, inserting this to our equation given to check if it a solution to y=ASinkt+BCoskt

4y''=-25y

For 4y''

4(-25/4ASin(±5/2t)-25/4BCos(±5/2t))

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