Question

Given that x^(2)+x=1, find the value of x^(3)+2x^(2)+2

Answer 1

Have you solve
\[x^2+x-1=0\] yet?

Answer 2

Solve for x

Answer 3

I'm supposed to modify one of the equations such that the other one can be substituted in

Answer 4

Solve the equation I mentioned and we will get to evaluating the expression for two values of x that we get

Answer 5

X=0.6180339887 OR X=-1.618033989

Answer 6

Can you approximate?

Answer 7

I mean are you allowed?

Answer 8

Well if that is the case, you could say that x = 1-x^2 and substitute that into the second equation but you will not get a numerical answer.

Answer 9

\[x=\frac{-1 \pm \sqrt{1-4(1)(-1)}}{2}=\frac{-1 \pm \sqrt{5}}{2}= \frac{-1 + \sqrt{5}}{2} \text{ or } \frac{-1 - \sqrt{5}}{2}\]

Answer 10

I'm not sure... I tried to substitute these two values into the second equation in the question but I got 3 & 3

Answer 11

How do I modify the 1st equation such that it can be substituted into 2nd equation?

Answer 12

\[(\frac{-1 +\sqrt{5}}{2})^3+2(\frac{-1+\sqrt{5}}{2})^2+2\]
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\[(\frac{-1 +\sqrt{5}}{2})^2=(\frac{-1 +\sqrt{5}}{2})(\frac{-1 + \sqrt{5}}{2})=\frac{1 -2 \sqrt{5}+5}{4}=\frac{6-2 \sqrt{5}}{4}\]
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\[(\frac{-1 + \sqrt{5}}{2})^3=(\frac{-1+\sqrt{5}}{2})^2(\frac{-1 + \sqrt{5}}{2})=(\frac{6 - 2 \sqrt{5}}{4})(\frac{-1 + \sqrt{5}}{2})\]
\[=\frac{-6 +3 \sqrt{5}-2(5)}{8}=\frac{-16+3 \sqrt{5}}{8}\]
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\[(\frac{-1 +\sqrt{5}}{2})^3+2(\frac{-1+\sqrt{5}}{2})^2+2\]
\[=\frac{-16 + 3 \sqrt{5}}{8}+2 \cdot \frac{6 - 2 \sqrt{5}}{4}+2\]
\[=\frac{-16 + 3 \sqrt{5}}{8}+\frac{12- 4 \sqrt{5}}{4}+2\]
\[=\frac{-16+3 \sqrt{5}+2(12-4 \sqrt{5})}{8}+2\]
\[=\frac{8-5 \sqrt{5}}{8}+2=\frac{8-5 \sqrt{5}+8(2)}{8}=\frac{24-5 \sqrt{5}}{8}\]
So I get not 3

Answer 13

but that is for one of the values

Answer 14

i leave the other one for you

Answer 15

Sorry, I don't really get your reply...

Answer 16

Well I found both of the values of x that solve x^2+x-1=0
Then I evaluated the expression x^3+2x^2+2 for one of the values I found (you can see what x I used above)