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Which equation is related to √x+10-1=x?
O x + 10 = x2 + x + 1
O x + 10 = x2 + 2x + 1
O x + 10 = x² +1
Ox+10=x²-1

Respuesta :

sarajeffries7

Answer:

  x + 10 = x² + 1

(third option listed)

Step-by-step explanation:

assuming that you meant: [tex]\sqrt{x+10-1}=x[/tex]  , because otherwise there would be no equivalent relationships,

(goal: isolate x on one side of the equation whilst having x on the other side of the equation also, like the equation in the question)*

   x + 10 = x² + 1

        -1            -1

   x + 10 - 1 = x²

  [tex]\sqrt{x+10-1} =\sqrt{x^2}[/tex]                  [find √ of both sides to isolate x]

[tex]\sqrt{x+10-1} =x[/tex]         (equation in problem)

       

{note: there's no reason to not simplify the equation to [tex]\sqrt{x+ 9[/tex], but the question leaves the equation that way, so I didn't simplify it either}

*I used this goal to decide which equations seemed about right, and then trying to test things out in my head

semsee45

Answer:

[tex]x+10=x^2+2x+1[/tex]

Step-by-step explanation:

Given equation:

[tex]\sqrt{x+10}-1=x[/tex]

Add 1 to both sides:

[tex]\implies \sqrt{x+10}-1+1=x+1[/tex]

[tex]\implies \sqrt{x+10}=x+1[/tex]

Square both sides:

[tex]\implies (\sqrt{x+10})^2=(x+1)^2[/tex]

[tex]\implies \sqrt{x+10}=(x+1)(x+1)[/tex]

Simplify:

[tex]\implies \sqrt{x+10}=x(x+1)+1(x+1)[/tex]

[tex]\implies \sqrt{x+10}=x^2+x+x+1[/tex]

[tex]\implies x+10=x^2+2x+1[/tex]

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