Question

question
solve for all possible values of x.
\\(\sqrt{6x + 46} = x + 9\\)
answer attempt 1 out of 2

Explanation:

Step1: Square both sides to eliminate square root

To solve the equation \(\sqrt{6x + 46}=x + 9\), we first square both sides. Squaring the left side \((\sqrt{6x + 46})^2\) gives \(6x+46\), and squaring the right side \((x + 9)^2\) expands to \(x^{2}+18x + 81\) (using the formula \((a + b)^2=a^{2}+2ab + b^{2}\) where \(a = x\) and \(b = 9\)). So we get the equation:
\(6x+46=x^{2}+18x + 81\)

Step2: Rearrange into quadratic equation

Subtract \(6x\) and \(46\) from both sides to set the equation to zero:
\(x^{2}+18x + 81-6x - 46=0\)
Simplify the like terms: \(x^{2}+(18x-6x)+(81 - 46)=0\), which simplifies to \(x^{2}+12x + 35 = 0\)

Step3: Factor the quadratic equation

We need to factor \(x^{2}+12x + 35\). We look for two numbers that multiply to \(35\) and add up to \(12\). The numbers \(5\) and \(7\) work because \(5\times7 = 35\) and \(5 + 7=12\). So the factored form is:
\((x + 5)(x + 7)=0\)

Step4: Solve for x using zero - product property

Using the zero - product property, if \(ab = 0\), then either \(a = 0\) or \(b = 0\). So we have two cases:
- Case 1: \(x+5 = 0\), then \(x=-5\)
- Case 2: \(x + 7=0\), then \(x=-7\)

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: \(\sqrt{6\times(-5)+46}=\sqrt{-30 + 46}=\sqrt{16}=4\)
Right side: \(-5 + 9 = 4\)
Since the left side equals the right side, \(x=-5\) is a valid solution.

- For \(x=-7\):
Left side: \(\sqrt{6\times(-7)+46}=\sqrt{-42 + 46}=\sqrt{4}=2\)
Right side: \(-7 + 9 = 2\)? Wait, no, \(-7+9 = 2\)? Wait, \(-7 + 9=2\), but the left side is \(\sqrt{6\times(-7)+46}=\sqrt{-42 + 46}=\sqrt{4} = 2\)? Wait, no, wait: \(\sqrt{6\times(-7)+46}=\sqrt{-42 + 46}=\sqrt{4}=2\), and the right side is \(-7 + 9 = 2\)? Wait, no, I made a mistake earlier. Wait, \(-7+9 = 2\), and the left side is \(2\), so is it valid? Wait, no, wait let's recalculate:

Wait, original equation: \(\sqrt{6x + 46}=x + 9\)

For \(x=-7\):

Left side: \(\sqrt{6\times(-7)+46}=\sqrt{-42 + 46}=\sqrt{4}=2\)

Right side: \(-7 + 9=2\). Wait, that seems valid? But wait, let's check again. Wait, maybe I made a mistake in factoring? Wait, no, the quadratic equation was \(x^{2}+12x + 35=(x + 5)(x + 7)\), that's correct. Wait, but let's check the original equation again for \(x=-7\):

\(\sqrt{6\times(-7)+46}=\sqrt{-42 + 46}=\sqrt{4}=2\), and \(x + 9=-7 + 9 = 2\). So both sides are equal? Wait, but earlier when I thought about the square root, the expression under the square root must be non - negative, and for \(x=-7\), \(6x + 46=6\times(-7)+46=-42 + 46 = 4\geq0\), and the square root of a non - negative number is non - negative, and the right side \(x + 9=-7 + 9 = 2\) is non - negative. Wait, but let's check the arithmetic again. Wait, maybe I made a mistake in the sign. Wait, no, \(-7+9 = 2\), and \(\sqrt{4}=2\), so \(x=-7\) also seems valid? But wait, let's check the original equation again. Wait, the original equation is \(\sqrt{6x + 46}=x + 9\). When \(x=-7\), left side is \(2\), right side is \(2\), so it is valid? But wait, let's check the quadratic solution again. Wait, when we squared the equation \(\sqrt{6x + 46}=x + 9\), we assumed that \(x + 9\geq0\) because the square root is non - negative. For \(x=-7\), \(x + 9=2\geq0\), and for \(x=-5\), \(x + 9 = 4\geq0\). Wait, but let's check the quadratic equation again. Wait, \(x^{2}+12x + 35 = 0\) has roots \(x=-5\) and \(x=-7\), and both satisfy the original equation? Wait, but let's re - evaluate:

Wait, for \(x=-7\):

\(\sqrt{6\times(-7)+46}=\sqrt{-42 + 46}=\sqrt{4}=2\)

\(x + 9=-7 + 9 = 2\). So \(2 = 2\), so it is valid. Wait, but maybe I made a mistake in the initial calculation. Wait, no, the steps seem correct. But wait, let's check with \(x=-5\):

\(\sqrt{6\times(-5)+46}=\sqrt{-30 + 46}=\sqrt{16}=4\)

\(x + 9=-5 + 9 = 4\), so \(4 = 4\), valid.

Wait, but the problem says "solve for all possible values of x". So both \(x=-5\) and \(x=-7\)? But wait, let's check the domain of the square root: \(6x+46\geq0\), so \(6x\geq - 46\), \(x\geq-\frac{46}{6}\approx - 7.666\). Both \(x=-5\) and \(x=-7\) are in the domain. And also, the right side \(x + 9\) must be non - negative because the square root is non - negative. For \(x=-7\), \(x + 9 = 2\geq0\), for \(x=-5\), \(x + 9 = 4\geq0\). So both are valid? Wait, but let's check the quadratic equation again. Wait, when we squared the equation \(\sqrt{6x + 46}=x + 9\), we get \(6x + 46=(x + 9)^2=x^{2}+18x + 81\), then \(x^{2}+12x + 35 = 0\), which factors to \((x + 5)(x + 7)=0\), so roots at \(x=-5\) and \(x=-7\). And both satisfy the original equation. Wait, but maybe I made a mistake in the check for \(x=-7\). Wait, no, the arithmetic is correct. So the solutions are \(x=-5\) and \(x=-7\)? Wait, but let's check with \(x=-7\) in the original equation:

Left side: \(\sqrt{6\times(-7)+46}=\sqrt{-42 + 46}=\sqrt{4}=2\)

Right side: \(-7 + 9 = 2\). So it works. With \(x=-5\):

Left side: \(\sqrt{6\times(-5)+46}=\sqrt{-30 + 46}=\sqrt{16}=4\)

Right side: \(-5 + 9 = 4\). So it works. So both are solutions? Wait, but maybe the problem has a typo? Or maybe I made a mistake. Wait, let's re - derive the equation.

Starting with \(\sqrt{6x + 46}=x + 9\)

Square both sides: \(6x + 46=x^{2}+18x + 81\)

Bring all terms to one side: \(x^{2}+12x + 35 = 0\)

Factor: \((x + 5)(x + 7)=0\), so \(x=-5\) or \(x=-7\)

Check \(x=-5\):

\(\sqrt{6*(-5)+46}=\sqrt{16}=4\), \(x + 9=4\), so 4 = 4: valid.

Check \(x=-7\):

\(\sqrt{6*(-7)+46}=\sqrt{4}=2\), \(x + 9=2\), so 2 = 2: valid.

So both are solutions. Wait, but maybe the original problem was written incorrectly? Or maybe I misread it. Wait, the original equation is \(\sqrt{6x + 46}=x + 9\). Yes, that's what is given. So the solutions are \(x=-5\) and \(x=-7\)? Wait, but let's check the value of the square root and the right side again. For \(x=-7\), the square root is 2, and the right side is 2, which is correct. For \(x=-5\), square root is 4, right side is 4, correct. So both are valid.

Wait, but maybe the problem expects only one solution? Wait, no, according to the calculations, both are valid.

Answer:

The solutions are \(x=-5\) and \(x=-7\)