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Problem #47 · September 12, 2026

Powers and roots

If x=20+20+20+…x = \sqrt{20 + \sqrt{20 + \sqrt{20 + \dots}}}, find the value of xx.
  1. A252\sqrt{5}
  2. B44
  3. C1010
  4. D2020
  5. E55

WORKED EXAMPLE

Solution

The value of the inner infinite root is exactly equal to xx itself, so x=20+xx = \sqrt{20 + x}. We square both sides: x2=20+xx^{2} = 20 + x, that is, x2−x−20=0x^{2} - x - 20 = 0. From this, (x−5)(x+4)=0(x-5)(x+4) = 0; since the root is positive, x=5x = 5. Check: 52=25=20+55^{2} = 25 = 20 + 5.
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