Question

solve for all possible values of x.\\(\sqrt{9x + 58} = x + 8\\)\
answer attempt 1 out of 2\
\\(\oplus\\) additional solution \\(\ominus\\) no solution\
\\(x = \square\\)\
submit answer

Explanation:

Step1: Square both sides to eliminate the square root

To solve the equation \(\sqrt{9x + 58}=x + 8\), we first square both sides. Squaring the left side \((\sqrt{9x + 58})^2\) gives \(9x+58\), and squaring the right side \((x + 8)^2\) using the formula \((a + b)^2=a^2+2ab + b^2\) (where \(a = x\) and \(b = 8\)) gives \(x^{2}+16x + 64\). So we get the equation:
\[9x+58=x^{2}+16x + 64\]

Step2: Rearrange into standard quadratic form

Subtract \(9x\) and \(58\) from both sides to set the equation to zero:
\[x^{2}+16x+64 - 9x - 58=0\]
Simplify the like terms: \(16x-9x = 7x\) and \(64 - 58=6\), so the equation becomes:
\[x^{2}+7x + 6=0\]

Step3: Factor the quadratic equation

We need to factor \(x^{2}+7x + 6\). We look for two numbers that multiply to \(6\) (the constant term) and add up to \(7\) (the coefficient of \(x\)). The numbers \(1\) and \(6\) work because \(1\times6 = 6\) and \(1 + 6=7\). So we can factor the quadratic as:
\[(x + 1)(x + 6)=0\]

Step4: Solve for \(x\) using the zero - product property

The zero - product property states that if \(ab = 0\), then either \(a = 0\) or \(b = 0\). So we set each factor equal to zero:
- If \(x+1 = 0\), then \(x=-1\).
- If \(x + 6=0\), then \(x=-6\).

Step5: Check for extraneous solutions

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

- For \(x=-1\):
Left side: \(\sqrt{9(-1)+58}=\sqrt{-9 + 58}=\sqrt{49}=7\)
Right side: \(-1 + 8=7\)
Since the left side equals the right side, \(x=-1\) is a valid solution.

- For \(x=-6\):
Left side: \(\sqrt{9(-6)+58}=\sqrt{-54 + 58}=\sqrt{4}=2\)
Right side: \(-6 + 8=2\)
Wait, wait, let's recalculate the left side for \(x = - 6\): \(9\times(-6)+58=-54 + 58 = 4\), \(\sqrt{4}=2\), and the right side is \(-6 + 8 = 2\). Wait, but earlier when we squared, we might have made a mistake? Wait no, let's check again. Wait, no, when \(x=-6\), original equation: \(\sqrt{9\times(-6)+58}=\sqrt{-54 + 58}=\sqrt{4}=2\), and \(x + 8=-6 + 8 = 2\). But wait, let's check the domain of the square root. The expression inside the square root \(9x + 58\) must be non - negative. For \(x=-6\), \(9\times(-6)+58=4\geq0\), and for \(x=-1\), \(9\times(-1)+58 = 49\geq0\). Wait, but let's check the original equation again for \(x=-6\): \(\sqrt{9x + 58}=\sqrt{9\times(-6)+58}=\sqrt{4}=2\), and \(x + 8=-6 + 8 = 2\). So both seem to work? Wait, no, wait I made a mistake in the factoring step? Wait, no, the quadratic was \(x^{2}+7x + 6=(x + 1)(x + 6)\), that's correct. Wait, but let's re - check the squaring step. Original equation: \(\sqrt{9x + 58}=x + 8\). For \(x=-6\), the right side is \(x + 8=2\), and the left side is \(\sqrt{9x + 58}=\sqrt{4}=2\). So \(x=-6\) also works? Wait, but when I first calculated, I thought maybe there was a mistake, but according to the calculations, both work. Wait, but let's check the original problem again. Wait, the original equation is \(\sqrt{9x + 58}=x + 8\). Let's check \(x=-6\) again:

Left: \(\sqrt{9*(-6)+58}=\sqrt{-54 + 58}=\sqrt{4}=2\)

Right: \(-6 + 8=2\). So it is valid. Wait, but maybe I made a mistake in the initial thought about extraneous solutions. But let's check the problem again. Wait, the user's problem is to solve for all possible values of \(x\). So both \(x=-1\) and \(x=-6\)? Wait, no, wait when \(x=-6\), let's check the domain of the square root: \(9x + 58\geq0\), \(9x\geq - 58\), \(x\geq-\frac{58}{9}\approx - 6.44\). Both \(x=-6\) (which is greater than \(-\frac{58}{9}\)) and \(x=-1\) (also greater than \(-\frac{58}{9}\)) are in the domain. Wait, but let's check the original equation again for \(x=-6\):

\(\sqrt{9*(-6)+58}=\sqrt{4}=2\), and \(x + 8=-6 + 8 = 2\). So it is a valid solution. Wait, but maybe I made a mistake in the factoring? Wait, \(x^{2}+7x + 6=(x + 1)(x + 6)\), that's correct. So the solutions are \(x=-1\) and \(x=-6\)? But the problem says "Answer Attempt 1 out of 2" and has an "Additional Solution" button. Wait, maybe I made a mistake in the check for \(x=-6\). Wait, let's re - evaluate the original equation for \(x=-6\):

The right - hand side is \(x + 8=-6 + 8 = 2\), and the left - hand side is \(\sqrt{9x + 58}=\sqrt{9*(-6)+58}=\sqrt{-54 + 58}=\sqrt{4}=2\). So it is a valid solution. Wait, but maybe the problem expects both solutions? But let's check the problem again. Wait, the original equation is \(\sqrt{9x + 58}=x + 8\). Let's solve it again:

Square both sides: \(9x + 58=x^{2}+16x + 64\)

\(x^{2}+7x + 6 = 0\)

\((x + 1)(x + 6)=0\)

Solutions \(x=-1\) and \(x=-6\). Both satisfy the original equation. But maybe the problem has a typo or I made a mistake? Wait, no, according to the calculations, both are valid. Wait, but let's check with \(x=-6\) in the original equation:

\(\sqrt{9*(-6)+58}=\sqrt{4}=2\), and \(x + 8=2\). So it works. So the solutions are \(x=-1\) and \(x=-6\). But the user's interface has an "Additional Solution" button, so maybe we need to list both. But let's check the problem statement again. The problem says "Solve for all possible values of \(x\)". So we have two solutions: \(x=-1\) and \(x=-6\). But wait, when I first calculated, I thought maybe \(x=-6\) was extraneous, but it's not. Wait, maybe I made a mistake in the sign when squaring? No, squaring both sides is a valid operation as long as we check the solutions. So both \(x=-1\) and \(x=-6\) are solutions. But let's check the original problem's interface: it has a box for \(x=\) and an "Additional Solution" button. So we can enter one solution and then add the other. But let's confirm with the calculations.

Wait, let's re - check \(x=-6\):

Left: \(\sqrt{9*(-6)+58}=\sqrt{-54 + 58}=\sqrt{4}=2\)

Right: \(-6 + 8=2\). Correct.

\(x=-1\):

Left: \(\sqrt{9*(-1)+58}=\sqrt{49}=7\)

Right: \(-1 + 8=7\). Correct.

So both are solutions. But maybe the problem expects us to list them. However, let's see the initial problem's "Answer Attempt 1 out of 2" and "Additional Solution" button. So we can first put \(x=-1\) (or \(x=-6\)) and then add the other. But according to the calculation, both are valid.

Answer:

The solutions are \(x=-1\) and \(x=-6\). If we consider the first solution (for the first attempt), we can write \(x=-1\) (or \(x=-6\), but \(x=-1\) is also valid).