Square Root Equations: Solve (√x+4)(√x-4)=0 and √x=4 for Largest X

To solve these equations, we'll follow these steps:

• Step 1: Solve $(\sqrt{x}+4)(\sqrt{x}-4)=0$
• Step 2: Solve $\sqrt{x}=4$
• Step 3: Compare both solutions to find the largest value of $x$

Let's work through each step:

Step 1: For the equation $(\sqrt{x}+4)(\sqrt{x}-4)=0$, note that it is in the form of a difference of squares: $a^2 - b^2 = (a-b)(a+b) = 0$. Here, $a = \sqrt{x}$ and $b = 4$, giving:

$a^2 - b^2 = (\sqrt{x})^2 - 4^2 = x - 16 = 0$

This simplifies to $x - 16 = 0$, meaning $x = 16$.

Step 2: For the equation $\sqrt{x} = 4$, squaring both sides yields:

$(\sqrt{x})^2 = 4^2$ $x = 16$

Step 3: Now that both equations result in $x = 16$, there is only one unique solution. Comparing the single value from each equation, $x = 16$ is the largest.

Therefore, the largest $x$ is $x = 16$.