Question

What is the value of \(\sqrt{42 + \sqrt{42 + \sqrt{42 + \sqrt{42 + \cdots}}}}∞\) ?

• A. -7
• B. -6
• C. 6
• D. 7
• E. 42

Answer 1

The Correct Option is D

To find the value of the expression \(\sqrt{42 + \sqrt{42 + \sqrt{42 + \cdots}}}\), let's denote the expression by \( x \). Hence, we have:

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

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

\( x^2 = 42 + x \).

Rearrange the equation to form a quadratic equation:

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

The quadratic equation is \( ax^2 + bx + c = 0 \) with \( a = 1 \), \( b = -1 \), and \( c = -42 \). We can solve it using the quadratic formula:

\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).

Substitute the values of \( a \), \( b \), and \( c \) into the formula:

\( x = \frac{1 \pm \sqrt{(-1)^2 - 4 \times 1 \times (-42)}}{2 \times 1} \).

Simplify under the square root:

\( x = \frac{1 \pm \sqrt{1 + 168}}{2} \).

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

The square root of 169 is 13, so we have:

\( x = \frac{1 \pm 13}{2} \).

This gives two possible solutions:

• \( x = \frac{1 + 13}{2} = \frac{14}{2} = 7 \).
• \( x = \frac{1 - 13}{2} = \frac{-12}{2} = -6 \).

Since \( x \) represents a length, which cannot be negative, the valid solution is \( x = 7 \).

Therefore, the value of the expression \(\sqrt{42 + \sqrt{42 + \sqrt{42 + \cdots}}}\) is 7.