Question

Find the real solutions of each equation. $$ x+\sqrt{x}=20 $$

Short answer

The real solution is \( x = 16 \).

Step-by-step solution

Isolate the Square Root Term

Subtract \(x\) from both sides of the equation to isolate the square root term: \[ x + \sqrt{x} = 20 \implies \sqrt{x} = 20 - x \]

Square Both Sides

Square both sides of the equation to eliminate the square root: \[ (\sqrt{x})^2 = (20 - x)^2 \implies x = (20 - x)^2 \]

Expand and Simplify

Expand the right-hand side and simplify: \[ x = 400 - 40x + x^2 \] Move all terms to one side to form a quadratic equation: \[ x^2 - 41x + 400 = 0 \]

Solve the Quadratic Equation

Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = -41\), and \(c = 400\): \[ x = \frac{41 \pm \sqrt{41^2 - 4 \times 1 \times 400}}{2 \times 1} \] Calculate the discriminant: \[ 41^2 - 4 \times 1 \times 400 = 1681 - 1600 = 81 \] So, the solutions are: \[ x = \frac{41 \pm 9}{2} \] Which gives: \[ x = 25 \] and \[ x = 16 \]

Verify the Solutions

Substitute both potential solutions back into the original equation to ensure they satisfy it: For \(x = 25\), \[25 + \sqrt{25} = 25 + 5 = 30 \text{ (Not a valid solution)} \] For \(x = 16\), \[16 + \sqrt{16} = 16 + 4 = 20 \text{ (Valid solution)} \]

Key concepts

Isolating Terms

When solving equations, one of the crucial steps is isolating the terms. This means you want to get the term with the variable you are solving for by itself on one side of the equation.
In the given problem, we start with the equation: \[ x + \sqrt{x} = 20 \]The goal is to isolate the square root term. By subtracting \(x\) from both sides, we achieve this: \[ \sqrt{x} = 20 - x \]
Isolating terms simplifies the equation and makes the next steps more manageable.

Quadratic Formula

The quadratic formula is a powerful tool used to solve quadratic equations of the form \(ax^2 + bx + c = 0\). The general form of the quadratic formula is: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]In the problem, after isolating and squaring the terms, we get a quadratic equation: \[ x^2 - 41x + 400 = 0 \]Here, \(a = 1\), \(b = -41\), and \(c = 400\). We plug these values into the quadratic formula: \[ x = \frac{41 \pm \sqrt{41^2 - 4 \times 1 \times 400}}{2 \times 1} \]The discriminant (the part under the square root) will help us determine if there are real solutions: \[ 41^2 - 4 \times 1 \times 400 = 1681 - 1600 = 81 \]Since the discriminant is positive, we have two real solutions. These solutions are: \[ x = \frac{41 + 9}{2} = 25 \] and \[ x = \frac{41 - 9}{2} = 16 \]

Real Solutions

Once we have potential solutions, it's essential to verify if they are real and valid.
We found two potential solutions from the quadratic formula: \(x = 25\) and \(x = 16\).
To verify, we substitute these back into the original equation to check if they work: \[ 25 + \sqrt{25} = 25 + 5 = 30 \quad \text{(Not a valid solution)} \] \[ 16 + \sqrt{16} = 16 + 4 = 20 \quad \text{(Valid solution)} \] Hence, the only **real solution** to the original equation \(x + \sqrt{x} = 20\) is \(x = 16\).