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Question 12·Hard·Nonlinear Equations in One Variable; Systems in Two Variables

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|x-3| = \sqrt{2x-1}

What is the solution set of the equation above?

For equations that mix absolute values and square roots, first impose the domain condition from the square root (set the inside $\ge 0$ ). Then, because both sides are nonnegative, square both sides to remove the absolute value and the root, simplify to a polynomial equation, and solve it (often with the quadratic formula). Always plug every algebraic solution back into the original equation to eliminate any extraneous solutions caused by squaring, and make sure your final answer includes all and only the values that actually work.

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Think about where the square root is defined

Before you manipulate the equation, ask: for which $x$ is $\sqrt{2x-1}$ defined? Turn that into an inequality and solve it.

Clear the absolute value and square root carefully

Both $|x-3|$ and $\sqrt{2x-1}$ are nonnegative. What operation can you do to both sides to remove the absolute value bars and the square root symbol at the same time?

Solve the resulting quadratic and then check solutions

After you square both sides and simplify, you get a quadratic equation. Use factoring (if possible) or the quadratic formula, but remember that squaring can create extra solutions—so you must test each solution in the original equation.

Desmos Guide

Graph both sides of the equation

In Desmos, enter y1 = abs(x-3) for the left-hand side and y2 = sqrt(2x-1) for the right-hand side. Remember that the square root is only defined for $x \ge 0.5$ , so focus on that region of the graph.

Locate the intersection points

Adjust the viewing window if needed so you can clearly see where the graphs of $y1$ and $y2$ cross. There should be two intersection points where the V-shaped graph of $|x-3|$ meets the curve $\sqrt{2x-1}$ .

Read off the x-coordinates of the intersections

Tap each intersection point; Desmos will display its coordinates. The $x$ -values of these intersection points are the solutions to $|x-3| = \sqrt{2x-1}$ , and together they form the solution set.

Step-by-step Explanation

Set the domain and remove the absolute value and square root

Because of the square root, we must have $2x-1 \ge 0$ , so $x \ge \tfrac12$ .

Both sides of the equation

|x-3| = \sqrt{2x-1}

are nonnegative, so we can safely square both sides:

(|x-3|)^2 = (\sqrt{2x-1})^2.

This gives

(x-3)^2 = 2x-1.

Expand and simplify:

(x-3)^2 = x^2 - 6x + 9,\\[0.7em] \Rightarrow x^2 - 6x + 9 = 2x - 1,\\[0.7em] \Rightarrow x^2 - 8x + 10 = 0.

Now we just need to solve the quadratic $x^2 - 8x + 10 = 0$ .

Set up the quadratic formula

For the quadratic $x^2 - 8x + 10 = 0$ , the coefficients are $a=1$ , $b=-8$ , and $c=10$ .

Use the quadratic formula

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

Substitute $a$ , $b$ , and $c$ :

x = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(1)(10)}}{2(1)} = \frac{8 \pm \sqrt{64 - 40}}{2}.

These are the two candidate solutions; next we simplify and check them in the original equation.

Simplify, check for extraneous solutions, and state the solution set

Simplify the expression from the quadratic formula:

64 - 40 = 24,\\[0.7em] \sqrt{24} = \sqrt{4\cdot 6} = 2\sqrt{6},\\[0.7em] \Rightarrow x = \frac{8 \pm 2\sqrt{6}}{2} = 4 \pm \sqrt{6}.

So the candidates are $x = 4 - \sqrt{6}$ and $x = 4 + \sqrt{6}$ .

Both are greater than $\tfrac12$ , so they are in the domain of $\sqrt{2x-1}$ . Check them in the original equation:

For $x = 4 + \sqrt{6}$ , $|x-3| = |1+\sqrt{6}| = 1+\sqrt{6}$ and

\sqrt{2x-1} = \sqrt{2(4+\sqrt{6})-1} = \sqrt{7+2\sqrt{6}} = 1+\sqrt{6}.

For $x = 4 - \sqrt{6}$ , $|x-3| = |1-\sqrt{6}| = \sqrt{6}-1$ and

\sqrt{2x-1} = \sqrt{2(4-\sqrt{6})-1} = \sqrt{7-2\sqrt{6}} = \sqrt{6}-1.

In each case, left and right sides match, so both values are true solutions. Therefore, the solution set is $\{4 - \sqrt{6},\; 4 + \sqrt{6}\}$ .

Question 12·Hard·Nonlinear Equations in One Variable; Systems in Two Variables

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Step-by-step Explanation

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Strategy

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Desmos Guide

Step-by-step Explanation