Problem 49 - Algebra and Trigonometry

Question

1–54 ? Find all real solutions of the equation. $$ \sqrt{\sqrt{x+5}+x}=5 $$

Step-by-Step Solution

Answer

There are no real solutions for the equation.

Step 1: Understand the Equation

The given equation is \( \sqrt{\sqrt{x+5}+x} = 5 \). The goal is to find all real values of \( x \) that satisfy this equation. We recognize that there is a square root inside another square root.

Step 2: Eliminate the Outer Square Root

To eliminate the outer square root, square both sides of the equation. This gives us \( \sqrt{x+5} + x = 25 \).

Step 3: Simplify the Resulting Equation

Subtract \( x \) from both sides to isolate the inner square root: \( \sqrt{x+5} = 25 - x \).

Step 4: Eliminate the Inner Square Root

Square both sides again to remove the square root: \( x + 5 = (25 - x)^2 \). This becomes a quadratic equation.

Step 5: Expand and Rearrange the Quadratic Equation

Expand \((25 - x)^2\) to obtain \( 625 - 50x + x^2 \). Thus, the equation becomes \( x + 5 = x^2 - 50x + 625 \). Rearrange to form \( x^2 - 51x + 620 = 0 \).

Step 6: Solve the Quadratic Equation

Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) to solve \( x^2 - 51x + 620 = 0 \), where \( a = 1 \), \( b = -51 \), and \( c = 620 \). Calculate the discriminant: \( (-51)^2 - 4 \times 1 \times 620 = 1 \). The roots are \( x = \frac{51 \pm 1}{2} \).

Step 7: Calculate the Solutions

The solutions from the quadratic formula are \( x = \frac{52}{2} = 26 \) and \( x = \frac{50}{2} = 25 \). These are the potential solutions.

Step 8: Verify Solutions

Check each solution in the original equation to ensure they do not result in extraneous solutions. For \( x = 26 \), substitute back into the original equation: \( \sqrt{\sqrt{26+5} + 26} = \sqrt{\sqrt{31}+26} \) which is not equal to 5, so \( x = 26 \) is not valid. For \( x = 25 \), substitute back: \( \sqrt{\sqrt{25+5} + 25} = \sqrt{\sqrt{30}+25} \) which also does not simplify to 5. Hence, after verification, no real solutions from these values.

Key Concepts

Quadratic equation

A quadratic equation is any equation that can be rearranged in the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, and \( a \) is not equal to zero. This equation is called "quadratic" because "quad" means "square," and the variable \( x \) is squared. Quadratics can have zero, one, or two real solutions.
To find these solutions, we often use methods like factoring, completing the square, or the quadratic formula, which is \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Here, the discriminant \( b^2 - 4ac \) determines the nature of the roots:

• If it is positive, there are two distinct real roots.
• If it is zero, there is one real root.
• If it is negative, there are no real roots.

In our task, solving the equation \( x^2 - 51x + 620 = 0 \) required using the quadratic formula because factoring was not straightforward.

Square root elimination

Square root elimination involves removing the radical (square root) by squaring both sides of an equation. This is a crucial step when solving equations that contain nested roots, like the one we had: \( \sqrt{\sqrt{x+5}+x} = 5 \).
By squaring both sides, we first simplified the problem from a nested root to a single square root: \( \sqrt{x+5} + x = 25 \).

Next, we applied squaring again to eliminate the remaining root:

• Squaring removes the direct square roots, simplifying our equation and converting it into a standard form that allows further algebraic manipulation.
• Each squaring step must be followed carefully, as it can introduce extraneous solutions, meaning solutions that arise from the manipulation but do not satisfy the original equation.

Squaring allowed us to turn a complex expression into a manageable quadratic equation from which we could find potential solutions.

Verification of solutions

Verification of solutions is a crucial step in the problem-solving process. Once we obtain potential solutions from solving an equation, particularly when squaring was involved, we must ensure these solutions satisfy the original equation.
This means substituting our solutions back into the original equation to check if they hold true:

• For \( x = 26 \), substituting back did not satisfy \( \sqrt{\sqrt{x+5}+x} = 5 \).
• Similarly, \( x = 25 \) also failed to satisfy the original equation.

After testing both potential solutions, we discovered neither were valid, highlighting why verification is necessary. Without verification, we may incorrectly assume a solution is valid when it results from an algebraic manipulation, such as squaring, that expanded possible values beyond those satisfying the original condition. Always include verification to confirm that the solutions are not extraneous and are truly real solutions of the equation.