Question

question
solve for all possible values of x.
\\(\sqrt{5x + 26} = x + 6\\)
answer attempt 1 out of 2
\\(\oplus\\) additional solution \\(\circleddash\\) no solution
\\(x = \square\\) submit answer

Explanation:

Step1: Square both sides to eliminate the square root

To solve the equation \(\sqrt{5x + 26}=x + 6\), we first square both sides of the equation. Squaring the left side \((\sqrt{5x + 26})^2\) gives \(5x+26\), and squaring the right side \((x + 6)^2\) using the formula \((a + b)^2=a^2+2ab + b^2\) (where \(a=x\) and \(b = 6\)) gives \(x^{2}+12x + 36\). So we get the equation:
\[5x+26=x^{2}+12x + 36\]

Step2: Rearrange the equation to standard quadratic form

Subtract \(5x\) and \(26\) from both sides of the equation \(5x+26=x^{2}+12x + 36\) to set the equation to zero:
\[x^{2}+12x+36-(5x + 26)=0\]
Simplify the left side:
\[x^{2}+12x + 36-5x-26=0\]
Combine like terms:
\[x^{2}+(12x-5x)+(36 - 26)=0\]
\[x^{2}+7x + 10=0\]

Step3: Factor the quadratic equation

We factor the quadratic equation \(x^{2}+7x + 10=0\). We need two numbers that multiply to \(10\) and add up to \(7\). The numbers \(2\) and \(5\) satisfy this, so we can factor the equation as:
\[(x + 2)(x+5)=0\]

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 for \((x + 2)(x + 5)=0\), we have:
Case 1: \(x+2=0\), then \(x=-2\)
Case 2: \(x + 5=0\), then \(x=-5\)

Step5: Check the solutions in the original equation

- For \(x=-2\):
Substitute \(x = - 2\) into the original equation \(\sqrt{5x+26}=x + 6\).
Left side: \(\sqrt{5\times(-2)+26}=\sqrt{-10 + 26}=\sqrt{16}=4\)
Right side: \(-2 + 6=4\)
Since the left side equals the right side, \(x=-2\) is a valid solution.

- For \(x=-5\):
Substitute \(x=-5\) into the original equation \(\sqrt{5x+26}=x + 6\).
Left side: \(\sqrt{5\times(-5)+26}=\sqrt{-25 + 26}=\sqrt{1}=1\)
Right side: \(-5+6 = 1\)? No, \(-5 + 6=1\) is wrong, \(-5+6 = 1\) is correct? Wait, \(-5+6 = 1\), and the left side is also \(1\)? Wait, no, \(-5+6=1\), and \(\sqrt{5\times(-5)+26}=\sqrt{-25 + 26}=\sqrt{1}=1\). Wait, I made a mistake earlier. Wait, \(-5 + 6=1\), and the left side is \(1\), so is \(x=-5\) a solution? Wait, no, wait: \(\sqrt{5x + 26}\) must be non - negative, and \(x + 6\) must also be non - negative because it is equal to a square root. For \(x=-5\), \(x + 6=-5 + 6 = 1\) (which is non - negative), and \(\sqrt{5\times(-5)+26}=\sqrt{1}=1\). Wait, but let's re - check:
Wait, when \(x=-5\), original equation: \(\sqrt{5\times(-5)+26}=\sqrt{-25 + 26}=\sqrt{1}=1\), and \(x + 6=-5 + 6 = 1\). So both sides are equal? Wait, I think I made a mistake in my initial check. Wait, no, let's recalculate:
Wait, \(5x+26\) when \(x=-5\): \(5\times(-5)+26=-25 + 26 = 1\), square root of \(1\) is \(1\). And \(x + 6=-5+6 = 1\). So \(x=-5\) also satisfies the equation? Wait, but let's check the domain of the square root. The expression under the square root \(5x + 26\geq0\), so \(5x\geq - 26\), \(x\geq-\frac{26}{5}=-5.2\). Both \(x=-2\) and \(x=-5\) are greater than or equal to \(-5.2\). And \(x + 6\) for \(x=-5\) is \(1\) (non - negative), for \(x=-2\) is \(4\) (non - negative). Wait, but when I first calculated, I thought there was a mistake, but actually:

For \(x=-5\):
\(\sqrt{5\times(-5)+26}=\sqrt{-25 + 26}=\sqrt{1}=1\)
\(x + 6=-5 + 6=1\)
So \(x=-5\) is also a solution? Wait, no, wait the original equation is \(\sqrt{5x + 26}=x + 6\). Let's check again. Wait, maybe I made a mistake in factoring? Wait, the quadratic equation was \(x^{2}+7x + 10 = 0\), factored as \((x + 2)(x + 5)=0\), solutions \(x=-2\) and \(x=-5\). And both satisfy the original equation? Wait, let's check with \(x=-5\):

Left side: \(\sqrt{5*(-5)+26}=\sqrt{-25 + 26}=\sqrt{1}=1\)

Right side: \(-5 + 6=1\)

Yes, it works. Wait, but earlier I thought there was a mistake, but actually, it works. Wait, but let's check the problem again. 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 \(x=-2\) and \(x=-5\) are solutions? Wait, no, wait:

Wait, \(\sqrt{5x+26}=x + 6\). The right - hand side \(x + 6\) must be non - negative because the square root of a number is non - negative. So \(x+6\geq0\), \(x\geq - 6\). Both \(x=-2\) and \(x=-5\) are greater than or equal to \(-6\). And when we substitute, both work. Wait, but let's check the quadratic equation again. We had \(\sqrt{5x + 26}=x + 6\), square both sides: \(5x+26=x^{2}+12x + 36\), then \(x^{2}+7x + 10 = 0\), factored as \((x + 2)(x + 5)=0\), solutions \(x=-2\) and \(x=-5\). And both satisfy the original equation. Wait, but maybe I made a mistake in the check for \(x=-5\) earlier. So the solutions are \(x=-2\) and \(x=-5\)? Wait, no, wait:

Wait, when \(x=-5\), \(x + 6=1\), and \(\sqrt{5x+26}=\sqrt{1}=1\), so it is correct. When \(x=-2\), \(x + 6 = 4\), and \(\sqrt{5x+26}=\sqrt{16}=4\), so it is correct. So both are solutions? But let's check the problem statement again. Wait, maybe I made a mistake in the quadratic equation. Let's re - do the squaring step:

\((\sqrt{5x + 26})^2=(x + 6)^2\)

\(5x+26=x^{2}+12x + 36\)

\(x^{2}+12x+36-5x - 26=0\)

\(x^{2}+7x + 10=0\)

Yes, that's correct. Factoring: \(x^{2}+7x + 10=(x + 2)(x + 5)\), correct. So the solutions are \(x=-2\) and \(x=-5\)? Wait, but let's check with \(x=-5\) in the original equation again:

Left: \(\sqrt{5*(-5)+26}=\sqrt{-25 + 26}=\sqrt{1}=1\)

Right: \(-5 + 6=1\). Yes, it's correct. So both \(x=-2\) and \(x=-5\) are solutions? Wait, but the problem says "solve for all possible values of x". But maybe I made a mistake. Wait, no, the square root function \(\sqrt{y}\) has a range \(y\geq0\), and the right - hand side \(x + 6\) must also be non - negative because \(\sqrt{5x+26}\geq0\). So \(x+6\geq0\) implies \(x\geq - 6\). Both \(x=-2\) and \(x=-5\) are greater than or equal to \(-6\), so they are valid.

Wait, but let's check the problem again. The original equation is \(\sqrt{5x + 26}=x + 6\). Let's graph the two functions \(y=\sqrt{5x + 26}\) and \(y=x + 6\). The square root function \(y=\sqrt{5x + 26}\) is defined for \(x\geq-\frac{26}{5}=-5.2\), and it is an increasing function. The line \(y=x + 6\) has a slope of \(1\) and a y - intercept of \(6\). The intersection points are at \(x=-2\) and \(x=-5\) (since when \(x=-5\), \(y = 1\) for both; when \(x=-2\), \(y = 4\) for both). So both are solutions.

But wait, maybe the problem expects only one solution? Wait, no, according to the calculations, both \(x=-2\) and \(x=-5\) are solutions. Wait, but let's check the arithmetic again for \(x=-5\):

\(5x+26=5*(-5)+26=-25 + 26 = 1\), \(\sqrt{1}=1\)

\(x + 6=-5+6 = 1\). So \(1 = 1\), correct.

For \(x=-2\):

\(5x+26=5*(-2)+26=-10 + 26 = 16\), \(\sqrt{16}=4\)

\(x + 6=-2+6 = 4\). So \(4 = 4\), correct.

So the solutions are \(x=-2\) and \(x=-5\). But wait, the problem's "Answer" section has a box for \(x=\) and options for "Additional Solution" and "No Solution". So maybe we need to list both solutions. But let's check the problem again. Wait, maybe I made a mistake in the sign when squaring? No, squaring both sides is a valid operation when we check the solutions at the end.

Answer:

The solutions are \(x=-2\) and \(x=-5\). If we consider the "Additional Solution" option, we can first write \(x=-2\) and then \(x=-5\) as an additional solution. But according to the calculation, both \(x=-2\) and \(x=-5\) are valid. So the possible values of \(x\) are \(-2\) and \(-5\).