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655-705 (Hard)|   Algebra|      
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Bunuel
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If x + 1 = x^2 and x > 0, then 2x^4 is: 22 Apr 2022, 03:15
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C
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65% (hard)

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64% (02:49) correct 36% (02:53) wrong based on 191 sessions

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If \(x + 1 = x^2\) and \(x > 0\), then \(2x^4\) is:


A. \(6 + 4√5\)

B. \(3 + 5√5\)

C. \(7 + 3√5\)

D. \(5 + 3√5\)

E. \(4 + 3√5\)




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C
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andreagonzalez2k
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If x + 1 = x^2 and x > 0, then 2x^4 is: 24 Apr 2022, 18:05
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Solving \(x^2-x-1=0\) the positive root is \(x=\frac{1+\sqrt{5}}{2}\)

Playing a little with the expression:
\(x^2=x+1\)
\(x^4=(x+1)^2=x^2+2x+1=x+1+2x+1=3x+2\)
\(2x^4=6x+4\)

Substituting:
\(6*\frac{1+\sqrt{5}}{2}+4=3+3\sqrt{5}+4=7+3\sqrt{5}\)

IMO C
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If x + 1 = x^2 and x > 0, then 2x^4 is: 24 Apr 2022, 05:40
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Bunuel
If \(x + 1 = x^2\) and \(x > 0\), then \(2x^4\) is:


A. \(6 + 4√5\)

B. \(3 + 5√5\)

C. \(7 + 3√5\)

D. \(5 + 3√5\)

E. \(4 + 3√5\)




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\(x^2-x-1=0…….x^2-x+(\frac{1}{2})^2-(\frac{1}{2})^2-1=0….(x-\frac{1}{2})^2-\frac{5}{4}=0\)

\((x-\frac{1}{2}-\frac{\sqrt{5}}{2}) (x-\frac{1}{2}+\frac{\sqrt{5}}{2})=0\)

As x>0, \(x=\frac{1}{2}+\frac{\sqrt{5}}{2} =\frac{1+\sqrt{5}}{2}\)


Thus, \(x^2=\frac{(1+5+2\sqrt{5})}{4}=\frac{3+\sqrt{5}}{2}\)

=> \(x^4=\frac{(9+5+6\sqrt{5})}{4}=\frac{7+3\sqrt{5}}{2}\)

And, \(2x^4=2*\frac{7+3\sqrt{5}}{2}=7+3\sqrt{5}\)
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If x + 1 = x^2 and x > 0, then 2x^4 is: 05 Aug 2022, 07:32
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chetan2u
Bunuel
If \(x + 1 = x^2\) and \(x > 0\), then \(2x^4\) is:


A. \(6 + 4√5\)

B. \(3 + 5√5\)

C. \(7 + 3√5\)

D. \(5 + 3√5\)

E. \(4 + 3√5\)








\(x^2-x-1=0…….x^2-x+(\frac{1}{2})^2-(\frac{1}{2})^2-1=0….(x-\frac{1}{2})^2-\frac{5}{4}=0\)

\((x-\frac{1}{2}-\frac{\sqrt{5}}{2}) (x-\frac{1}{2}+\frac{\sqrt{5}}{2})=0\)

As x>0, \(x=\frac{1}{2}+\frac{\sqrt{5}}{2} =\frac{1+\sqrt{5}}{2}\)


Thus, \(x^2=\frac{(1+5+2\sqrt{5})}{4}=\frac{3+\sqrt{5}}{2}\)

=> \(x^4=\frac{(9+5+6\sqrt{5})}{4}=\frac{7+3\sqrt{5}}{2}\)

And, \(2x^4=2*\frac{7+3\sqrt{5}}{2}=7+3\sqrt{5}\)
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could you elbaorate or how you get (1/2)^2? I am following the rest, but I am hung-up at the start. explaining that would really help me out.
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If x + 1 = x^2 and x > 0, then 2x^4 is: 02 Jul 2024, 07:04
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The most loved sequence you’ll find in the nature (from sunflower seed to cyclones to the the structures of galaxies). x^2=x+1 or T_n=T_(n-1)+T_(n-2).


I was first introduced this concept while solving advanced level problem on Limits of calculus.

If you know the solution of the quadratic is the golden ratio of 1.6 .

Golden Ratio is one of the most common mathematical ratios in nature. We see this ratio everywhere from majestic landscapes like the Pyramids of Giza and the Mona Lisa to modern-day logos such as Twitter and Pepsi. Golden ratios are unique because of their golden proportion.

2x^4=2*(1.6)^4=close to 13.


Or x^4=3x+2 giving C as ans.

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If x + 1 = x^2 and x > 0, then 2x^4 is: 24 Sep 2025, 14:36
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