Question

Pls Help, Max Points

Answer 1

`\\ \rm\dashrightarrow √(6-x)=√(x+7)-1`

• Square both sides

`\\ \rm\dashrightarrow 6-x=x+7-2(√(x+7)+1`

`\\ \rm\dashrightarrow -2x-2=-2√(x+7)`

`\\ \rm\dashrightarrow -2(x+1)=-2√(x+7)`

`\\ \rm\dashrightarrow x+1=√(x+7)`

`\\ \rm\dashrightarrow x^2+2x+1=x+7`

`\\ \rm\dashrightarrow x^2+x-6=0`

`\\ \rm\dashrightarrow x^2+3x-2x-6=0`

`\\ \rm\dashrightarrow x(x+3)-2(x+3)=0`

`\\ \rm\dashrightarrow (x+3)(x-2)=0`

• x=-3,2

Putting 2 in Equation

• √6-2=2
• √9-(1)
• 3-1
• 2

Verified

Putting -3

• √6-(-3)=√(-3+7)-1
• √6+3=√4-(1)
• √9=2-1
• 3=1

Incorrect

Only x=2 is the solution

Answer 2

x = 2

Explanation:

`\large \begin{aligned}√(6-x) & =√(x+7)-1\\\\(√(6-x))^2 & =(√(x+7)-1)^2\\\\6-x & = x -2√(x+7)+8\\\\-2(x+1) & = -2√(x+7)\\\\(x+1) & = √(x+7)\\\\(x+1)^2 & = (√(x+7))^2\\\\x^2+2x+1 & =x+7\\\\x^2+x-6 & =0\\\\(x+3)(x-2) & =0\\\\\implies x & = -3, 2\end{aligned}`

Substituting found values of x into the original equation to verify:

`\large \begin{aligned}x=-3 \implies √(6-(-3)) & = √(-3+7)-1\\3 & \\eq 1 \rightarrow \textsf{incorrect}\\\\x=2 \implies √(6-2) & = √(2+7)-1\\2 & = 2 \rightarrow \textsf{correct}\end{aligned}`

Therefore, the solution is x = 2 only