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Li Juan solves the equation below by first squaring both sides of the equation. [tex]\sqrt{3-2w}=w+6[/tex]
What extraneous solution does Li Juan obtain?


Sagot :

Answer:

w = -11

Step-by-step explanation:

[tex]\sqrt{3 - 2w} = w + 6[/tex]

[tex](\sqrt{3 - 2w})^2 = (w + 6)^2[/tex]

[tex] 3 - 2w = w^2 + 12w + 36 [/tex]

[tex] w^2 + 14w + 33 = 0 [/tex]

[tex] (w + 11)(w + 3) = 0 [/tex]

[tex] w + 11 = 0 [/tex]   or   [tex] w + 3 = 0 [/tex]

[tex] w = -11 [/tex]   or   [tex] w = -3 [/tex]

When you square both sides of an equation, you must check all solutions for extraneous solutions.

Check w = -11.

[tex]\sqrt{3 - 2w} = w + 6[/tex]

[tex] \sqrt{3 - 2(-11)} = -11 + 6 [/tex]

[tex] \sqrt{3 + 22} = -5 [/tex]

[tex] \sqrt{25} = -5 [/tex]

[tex] 5 = -5 [/tex]

This is a false statement, so the solution w = -11 is extraneous since it does not satisfy the original equation.

Check w = -3.

[tex]\sqrt{3 - 2w} = w + 6[/tex]

[tex] \sqrt{3 - 2(-3)} = -3 + 6 [/tex]

[tex] \sqrt{3 + 6} = 3 [/tex]

[tex] \sqrt{9} = 3 [/tex]

[tex] 3 = 3 [/tex]

This is a true statement, so the solution w = -3 is valid.

Answer: w = -11