gesianeferreira
Answered

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Solve the equation. (Enter your answers as a comma-separated list. If there is no solution, enter NO SOLU X = 6 X = -x + 6 = x - 6 Identify any extraneous solution. (If there is no extraneous solution, enter NO SOLUTION.) 2​

Solve The Equation Enter Your Answers As A Commaseparated List If There Is No Solution Enter NO SOLU X 6 X X 6 X 6 Identify Any Extraneous Solution If There Is class=

Sagot :

LammettHash

Taking squares on both sides leads to

[tex]\sqrt{-x + 6} = x - 6[/tex]

[tex]\left(\sqrt{-x + 6}\right)^2 = (x - 6)^2[/tex]

[tex]-x + 6 = x^2 - 12x + 36[/tex]

[tex]x^2 - 11x + 30 = 0[/tex]

[tex](x - 5) (x - 6) = 0[/tex]

Solving for [tex]x[/tex], we get

[tex]x - 5 = 0 \text{ or } x - 6 = 0[/tex]

or

[tex]x = 5 \text{ or } x = 6[/tex].

Evaluating both sides of the starting equation at these solutions, we have

[tex]\sqrt{-6 + 6} = 6 - 6 \implies 0 = 0[/tex]

which is true, so [tex]\boxed{x=6}[/tex] is a valid solution. However,

[tex]\sqrt{-5 + 6} = 5 - 6 \implies \sqrt1 = -1 \implies 1 = -1[/tex]

which is not true, so [tex]\boxed{x=5}[/tex] is an extraneous solution.

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