Question

question
solve for all possible values of x.
\\(\sqrt{5x - 21} = x - 3\\)
answer attempt 1 out of 2
\\(\oplus\\) additional solution \\(\ominus\\) no solution
\\(x = \square\\)
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Explanation:

Step1: Square both sides to eliminate square root

To solve the equation \(\sqrt{5x - 21}=x - 3\), we first square both sides. Squaring the left side gives \(5x - 21\), and squaring the right side gives \((x - 3)^2=x^{2}-6x + 9\). So the equation becomes:

$$5x-21=x^{2}-6x + 9$$

Step2: Rearrange into quadratic equation

Rearrange the equation to standard quadratic form \(ax^{2}+bx + c = 0\). Subtract \(5x\) and add \(21\) to both sides:

$$x^{2}-6x + 9-5x + 21 = 0$$

Simplify the left side:

$$x^{2}-11x + 30 = 0$$

Step3: Factor the quadratic equation

Factor the quadratic \(x^{2}-11x + 30\). We need two numbers that multiply to \(30\) and add to \(- 11\). The numbers are \(-5\) and \(-6\), so:

$$(x - 5)(x - 6)=0$$

Step4: Solve for x

Set each factor equal to zero:

• For \(x - 5=0\), we get \(x = 5\)
• For \(x - 6=0\), we get \(x = 6\)

Step5: Check for extraneous solutions

We need to check these solutions in the original equation because squaring both sides can introduce extraneous solutions.

• For \(x = 5\): Left side \(\sqrt{5(5)-21}=\sqrt{25 - 21}=\sqrt{4}=2\), Right side \(5 - 3 = 2\). So \(x = 5\) is a valid solution.
• For \(x = 6\): Left side \(\sqrt{5(6)-21}=\sqrt{30 - 21}=\sqrt{9}=3\), Right side \(6 - 3 = 3\). So \(x = 6\) is a valid solution.

Answer:

\(x = 5\) or \(x = 6\) (If we consider the original problem's "Additional Solution" option, both \(5\) and \(6\) are solutions. But if we need to list them, the possible values are \(5\) and \(6\))