Respuesta :

semsee45

Answer:

x = 2

Step-by-step explanation:

[tex]\large \begin{aligned}\sqrt{6-x} & =\sqrt{x+7}-1\\\\(\sqrt{6-x})^2 & =(\sqrt{x+7}-1)^2\\\\6-x & = x -2\sqrt{x+7}+8\\\\-2(x+1) & = -2\sqrt{x+7}\\\\(x+1) & = \sqrt{x+7}\\\\(x+1)^2 & = (\sqrt{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}[/tex]

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

[tex]\large \begin{aligned}x=-3 \implies \sqrt{6-(-3)} & = \sqrt{-3+7}-1\\3 & \neq 1 \rightarrow \textsf{incorrect}\\\\x=2 \implies \sqrt{6-2} & = \sqrt{2+7}-1\\2 & = 2 \rightarrow \textsf{correct}\end{aligned}[/tex]

Therefore, the solution is x = 2 only

MisterBrainly

[tex]\\ \rm\dashrightarrow \sqrt{6-x}=\sqrt{x+7}-1[/tex]

  • Square both sides

[tex]\\ \rm\dashrightarrow 6-x=x+7-2(\sqrt{x+7}+1[/tex]

[tex]\\ \rm\dashrightarrow -2x-2=-2\sqrt{x+7}[/tex]

[tex]\\ \rm\dashrightarrow -2(x+1)=-2\sqrt{x+7}[/tex]

[tex]\\ \rm\dashrightarrow x+1=\sqrt{x+7}[/tex]

[tex]\\ \rm\dashrightarrow x^2+2x+1=x+7[/tex]

[tex]\\ \rm\dashrightarrow x^2+x-6=0[/tex]

[tex]\\ \rm\dashrightarrow x^2+3x-2x-6=0[/tex]

[tex]\\ \rm\dashrightarrow x(x+3)-2(x+3)=0[/tex]

[tex]\\ \rm\dashrightarrow (x+3)(x-2)=0[/tex]

  • 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

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