Question

What is the value of \(\sqrt{42+\sqrt{42+\sqrt{42+\sqrt{42+\sqrt{42+...\infin}}}}}\) ?

• A. -7
• B. -6
• C. 6
• D. 7
• E. 42

Answer 1

The Correct Option is D

To find the value of \(\sqrt{42+\sqrt{42+\sqrt{42+\sqrt{42+\sqrt{42+\ldots}}}}}\), we can denote this expression as \( x \). Therefore, we have:

\( x = \sqrt{42 + \sqrt{42 + \sqrt{42 + \sqrt{42 + \ldots}}}} \)

Since the structure is infinite and repeating, the inner expression under the square root can also be represented as \( x \). This leads to the equation:

\( x = \sqrt{42 + x} \)

To solve for \( x \), square both sides to eliminate the square root:

\( x^2 = 42 + x \)

Rearrange the equation to form a quadratic equation:

\( x^2 - x - 42 = 0 \)

Next, solve this quadratic equation using the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -1 \), and \( c = -42 \).

Calculate the discriminant:

\( b^2 - 4ac = (-1)^2 - 4(1)(-42) = 1 + 168 = 169 \)

Now substitute back into the quadratic formula:

\( x = \frac{-(-1) \pm \sqrt{169}}{2(1)} = \frac{1 \pm 13}{2} \)

This yields two potential solutions for \( x \):

• \( x = \frac{1 + 13}{2} = 7 \)
• \( x = \frac{1 - 13}{2} = -6 \)

Since \( x \) represents a square root value, it must be non-negative. Therefore, the only viable solution is:

x = 7

Thus, the value of \(\sqrt{42+\sqrt{42+\sqrt{42+\sqrt{42+\ldots}}}}\) is 7.