Question

1. \sqrt{(x + 1)^2} = 75

Explanation:

Step1: Simplify the square root

The square root of a square, \(\sqrt{(x + 1)^2}\), simplifies to the absolute value of the expression inside, so \(\sqrt{(x + 1)^2}=|x + 1|\). But if we assume we are dealing with real numbers and the equation is \(\sqrt{(x + 1)^2}=75\) (correcting the typo in the original, assuming it's an equal sign), then \(|x + 1| = 75\). However, if we consider the principal square root (and if \(x+1\) is a real number, and we can also think of \(\sqrt{(x + 1)^2}=|x + 1|\), but if we are in a context where we can drop the absolute value by considering the domain or if we assume \(x+1\) is non - negative, we can also note that \(\sqrt{a^2}=a\) when \(a\geq0\). Let's first consider the equation as \(\sqrt{(x + 1)^2}=75\).

Step2: Solve for \(x\)

Case 1: If \(x + 1\geq0\), then \(\sqrt{(x + 1)^2}=x + 1\). So we have the equation \(x+1 = 75\). Subtract 1 from both sides: \(x=75 - 1=74\).

Case 2: If \(x + 1<0\), then \(\sqrt{(x + 1)^2}=-(x + 1)\). So we have the equation \(-(x + 1)=75\). Multiply both sides by - 1: \(x + 1=-75\). Subtract 1 from both sides: \(x=-75 - 1=-76\).

Answer:

The solutions for \(x\) are \(x = 74\) or \(x=-76\)