Examkin

Question

Difficulty: HardSurds and Indices

If p=12+12+12+…p = \sqrt{12 + \sqrt{12 + \sqrt{12 + \dots}}} and q=12−12−12−…q = \sqrt{12 - \sqrt{12 - \sqrt{12 - \dots}}}, such that 2p+q−z=1642^{p+q-z} = \frac{1}{64}, what is the value of zz?

  1. A
    11
  2. B
    55
  3. 1313Answer
  4. D
    −1-1

Answer

The value of zz is 1313.
Solving p=12+pp = \sqrt{12 + p} yields the quadratic equation p2−p−12=0p^2 - p - 12 = 0, giving p=4p = 4. Solving q=12−qq = \sqrt{12 - q} yields q2+q−12=0q^2 + q - 12 = 0, giving q=3q = 3. Adding pp and qq gives p+q=7p + q = 7. Substituting this into 2p+q−z=1642^{p+q-z} = \frac{1}{64} produces 27−z=2−62^{7-z} = 2^{-6}. Equating the indices gives 7−z=−67 - z = -6, which solves to z=13z = 13.

Step-by-Step Solution

1
Evaluate the first infinite nested surd pp
p=4p = 4
Let p=12+pp = \sqrt{12 + p}. Squaring both sides yields p2=12+p  ⟹  p2−p−12=0p^2 = 12 + p \implies p^2 - p - 12 = 0. Factoring gives (p−4)(p+3)=0(p-4)(p+3) = 0. Since p>0p > 0, p=4p = 4.
2
Evaluate the second infinite nested surd qq
q=3q = 3
Let q=12−qq = \sqrt{12 - q}. Squaring both sides yields q2=12−q  ⟹  q2+q−12=0q^2 = 12 - q \implies q^2 + q - 12 = 0. Factoring gives (q+4)(q−3)=0(q+4)(q-3) = 0. Since q>0q > 0, q=3q = 3.
3
Substitute pp and qq into the exponent expression
p+q=7p + q = 7, giving 27−z=1642^{7-z} = \frac{1}{64}
Summing pp and qq gives 4+3=74 + 3 = 7.
4
Solve the exponential equation for zz
z=13z = 13
Express 164\frac{1}{64} as a base 22 power: 164=2−6\frac{1}{64} = 2^{-6}. Equating exponents gives 7−z=−6  ⟹  z=7+6=137 - z = -6 \implies z = 7 + 6 = 13.

Key Concept

Evaluation of infinite nested surds and solving exponential equations using laws of indices
Estimated Time:2m 0s
Rate this question

Topics

Surds and Indices

More questions from this topic