Question

Solve:
√ (x²- 14 + 49) = x - 7

Answer 1

$Answer: \\ \sqrt{({x}^{2} - 14x + 49)} = x - 7 \\ \Leftrightarrow \sqrt{ {(x - 7)}^{2} } = x - 7 \\ \Leftrightarrow |x - 7| = x - 7 \\ \Leftrightarrow x - 7 = x - 7 \: \vee \: x - 7 = 7 - x \\ \Leftrightarrow x \in R \: \vee \:x = 7 \\ \Rightarrow x\in[7,\infty)$

Answer 2

Answer:

$\large\boxed{x\geq7\to x\in[7,\ \infty)}$

Step-by-step explanation:

$\sqrt{x^2-14x+49}=x-7\\\\\underbrace{\sqrt{x^2-2(x)(7)+7^2}}_{(*)}=x-7\qquad\text{use}\ (a-b)^2=a^2-2ab+b^2\\\\\sqrt{(x-7)^2}=x-7\qquad\text{use}\ \sqrt{a^2}=|a|\\\\|x-7|=x-7\qquad\text{use the de}\text{finition of an absolute value}\\\\|x-7|=\left\{\begin{array}{ccc}x-7&for&x\geq7\\-(x-7)&for&x<7\end{array}\right$

$(1)\ x<7\to x\in(-\infty,\ 7)\\\\-(x-7)=x-7\\-x-(-7)=x-7\\-x+7=x-7\qquad\text{subtract 7 from both sides}\\-x=x-14\qquad\text{subtract x from both sides}\\-2x=-14\qquad\text{divide both sides by (-2)}\\x=7\notin(-\infty,\ 7)\\\\NO\ SOLUTION$

$(2)\ x\geq7\to x\in[7,\ \infty)\\\\x-7=x-7\qquad\text{subtract x from both sides}\\-7=-7\qquad TRUE\\\\x\in[7,\ \infty)$