Question

ossible values of x.
(sqrt{7x + 18} = x + 4)
attempt 1 out of 2
no solution
solution

Explanation:

Step1: Square both sides to eliminate the square root

To solve the equation \(\sqrt{7x + 18}=x + 4\), we first square both sides. Squaring the left side gives us \(7x + 18\) (since \((\sqrt{a})^2=a\) for \(a\geq0\)), and squaring the right side \((x + 4)^2\) expands to \(x^2+8x + 16\) using the formula \((a + b)^2=a^2+2ab + b^2\) where \(a=x\) and \(b = 4\). So we get the equation:
\[7x+18=x^{2}+8x + 16\]

Step2: Rearrange into standard quadratic form

Subtract \(7x\) and \(18\) from both sides to set the equation to zero:
\[x^{2}+8x + 16-7x - 18=0\]
Simplify the like - terms:
\[x^{2}+(8x-7x)+(16 - 18)=0\]
\[x^{2}+x-2=0\]

Step3: Factor the quadratic equation

We need to find two numbers that multiply to \(- 2\) and add up to \(1\). The numbers are \(2\) and \(-1\). So we can factor the quadratic equation as:
\((x + 2)(x-1)=0\)

Step4: Solve for x using the zero - product property

If \((x + 2)(x - 1)=0\), then either \(x+2=0\) or \(x - 1=0\).
- For \(x+2=0\), we get \(x=-2\).
- For \(x - 1=0\), we get \(x = 1\).

Step5: Check for extraneous solutions

We need to check these solutions in the original equation because squaring both sides can introduce extraneous solutions.
- For \(x=-2\):
Left side: \(\sqrt{7\times(-2)+18}=\sqrt{-14 + 18}=\sqrt{4}=2\)
Right side: \(-2 + 4=2\)
So \(x=-2\) is a valid solution? Wait, wait, let's re - check. Wait, when \(x=-2\), the right side is \(x + 4=-2 + 4 = 2\), and the left side is \(\sqrt{7\times(-2)+18}=\sqrt{-14 + 18}=\sqrt{4}=2\). But wait, we also need to check the domain of the square root. The expression inside the square root \(7x+18\) must be non - negative. For \(x=-2\), \(7\times(-2)+18=-14 + 18 = 4\geq0\). But wait, let's check \(x = 1\):
Left side: \(\sqrt{7\times1+18}=\sqrt{7 + 18}=\sqrt{25}=5\)
Right side: \(1+4=5\). And for \(x = 1\), \(7\times1+18=25\geq0\). Wait, but when we check \(x=-2\) in the original equation \(\sqrt{7x + 18}=x + 4\), \(\sqrt{7\times(-2)+18}=\sqrt{4}=2\), and \(x + 4=-2 + 4=2\). But let's see the domain of the square root function \(y=\sqrt{7x + 18}\), the domain is \(7x+18\geq0\Rightarrow x\geq-\frac{18}{7}\approx - 2.571\). \(x=-2\) is in the domain. But wait, let's check the original equation again. Wait, maybe I made a mistake. Wait, when \(x=-2\), the right side is \(x + 4=2\), left side is \(\sqrt{4}=2\). But let's check the quadratic solution again. Wait, the quadratic equation was \(x^{2}+x - 2=0\), factored as \((x + 2)(x - 1)=0\), so roots \(x=-2\) and \(x = 1\). But let's check \(x=-2\) in the original equation: \(\sqrt{7\times(-2)+18}=\sqrt{4}=2\), and \(x + 4=2\), so it works? Wait, but let's see the graph of \(y=\sqrt{7x + 18}\) and \(y=x + 4\). The square root function \(y=\sqrt{7x+18}\) has a domain \(x\geq-\frac{18}{7}\), and it is an increasing function. The line \(y=x + 4\) has a slope of \(1\) and y - intercept \(4\). Let's find the intersection points. We found \(x=-2\) and \(x = 1\). But when \(x=-2\), the value of \(x + 4=2\), and \(\sqrt{7\times(-2)+18}=2\). When \(x = 1\), \(x + 4=5\) and \(\sqrt{25}=5\). But wait, is there a mistake here? Wait, let's re - do the squaring step. \((\sqrt{7x + 18})^2=7x + 18\) and \((x + 4)^2=x^{2}+8x + 16\). So \(7x+18=x^{2}+8x + 16\) simplifies to \(x^{2}+x - 2=0\), which factors to \((x + 2)(x - 1)=0\). So both solutions are valid? But wait, let's check with \(x=-2\): \(\sqrt{7\times(-2)+18}=\sqrt{4}=2\), \(x + 4=2\), so it's a solution. With \(x = 1\): \(\sqrt{25}=5\), \(x + 4=5\), so it's a solution. But wait, maybe the problem is that when we square both sides, we can introduce extraneous solutions, but in this case, both seem to work. Wait, but let's check the original problem again. Wait, the user's problem is to find the possible values of \(x\) for \(\sqrt{7x + 18}=x + 4\). So both \(x=-2\) and \(x = 1\)? But wait, let's check the domain of the square root: \(7x+18\geq0\Rightarrow x\geq-\frac{18}{7}\approx - 2.57\). Both \(x=-2\) and \(x = 1\) are in the domain. But wait, when I first thought, maybe I made a mistake. Wait, let's compute \(x=-2\): \(\sqrt{7*(-2)+18}=\sqrt{4}=2\), and \(x + 4=2\), so it's correct. \(x = 1\): \(\sqrt{25}=5\), \(x + 4=5\), correct. But wait, maybe the problem is that when \(x=-2\), the right - hand side \(x + 4 = 2\), and the left - hand side is \(2\), so it's a valid solution. But let's check the quadratic equation again. The equation after squaring is \(7x+18=x^{2}+8x + 16\), which is \(x^{2}+x - 2=0\), solutions \(x=-2\) and \(x = 1\). Both satisfy the original equation. Wait, but maybe I made a mistake in the domain? For \(x=-2\), \(7x+18=4\geq0\), so it's in the domain. So both \(x=-2\) and \(x = 1\) are solutions? But wait, let's check with \(x=-2\) in the original equation: \(\sqrt{7*(-2)+18}=\sqrt{4}=2\), and \(x + 4=2\), so it's correct. With \(x = 1\), \(\sqrt{25}=5\) and \(x + 4=5\), correct. So the solutions are \(x=-2\) and \(x = 1\)? Wait, but maybe the problem is that when we square both sides, we can have extraneous solutions, but in this case, both are valid. Wait, no, wait a second. Let's check the original equation again. The equation is \(\sqrt{7x + 18}=x + 4\). The square root function always returns a non - negative value, so the right - hand side \(x + 4\) must also be non - negative because the left - hand side is non - negative. So \(x+4\geq0\Rightarrow x\geq - 4\). But we already have from the square root that \(x\geq-\frac{18}{7}\approx - 2.57\). So for \(x=-2\), \(x + 4=2\geq0\), so it's okay. For \(x = 1\), \(x + 4=5\geq0\), okay. So both solutions are valid. Wait, but maybe I made a mistake in the factoring. Wait, \(x^{2}+x - 2\) factors as \((x + 2)(x - 1)\) because \(2\times(-1)=-2\) and \(2+(-1)=1\). Yes, that's correct. So the solutions are \(x=-2\) and \(x = 1\). But wait, let's check with \(x=-2\) again. \(\sqrt{7*(-2)+18}=\sqrt{4}=2\), and \(x + 4=2\), so it's correct. So the possible values of \(x\) are \(x=-2\) and \(x = 1\). Wait, but maybe the problem expects only \(x = 1\)? Wait, no, let's check again. Wait, when \(x=-2\), the original equation holds. So both are solutions.

Wait, maybe I made a mistake in the domain of the right - hand side. The left - hand side is a square root, so it's non - negative, so the right - hand side \(x + 4\) must also be non - negative. So \(x+4\geq0\Rightarrow x\geq - 4\). And the domain of the square root is \(x\geq-\frac{18}{7}\approx - 2.57\). So \(x=-2\) is in both domains. So both solutions are valid.

Answer:

The possible values of \(x\) are \(x = 1\) and \(x=-2\). But wait, when we check \(x=-2\) in the original equation, it works. However, let's re - check the calculation for \(x=-2\):

Left side: \(\sqrt{7\times(-2)+18}=\sqrt{-14 + 18}=\sqrt{4}=2\)

Right side: \(-2 + 4=2\)

So it works. And for \(x = 1\):

Left side: \(\sqrt{7\times1+18}=\sqrt{25}=5\)

Right side: \(1 + 4=5\)

So it works. So the solutions are \(x=-2\) and \(x = 1\). But maybe the problem has a typo or I misread it. Wait, the original equation is \(\sqrt{7x + 18}=x + 4\). So the solutions are \(x = 1\) and \(x=-2\).