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Li Juan solves the equation below by first squaring both sides of the equation. sqrt(3-2w)=w+6 What extraneous solution does Li Juan obtain? w=

Li Juan solves the equation below by first squaring both sides of the equation.\newline3−2w=w+6 \sqrt{3-2 w}=w+6 \newlineWhat extraneous solution does Li \mathrm{Li} Juan obtain?\newlinew= w=

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Q. Li Juan solves the equation below by first squaring both sides of the equation.\newline3−2w=w+6 \sqrt{3-2 w}=w+6 \newlineWhat extraneous solution does Li \mathrm{Li} Juan obtain?\newlinew= w=
  1. Square both sides: Square both sides of the equation to eliminate the square root.\newline(3−2w)2=(w+6)2(\sqrt{3 - 2w})^2 = (w + 6)^2\newline3−2w=w2+12w+363 - 2w = w^2 + 12w + 36
  2. Rearrange and find values: Rearrange the equation to set it to zero and find the values of ww.0=w2+12w+36−(3−2w)0 = w^2 + 12w + 36 - (3 - 2w)0=w2+12w+36−3+2w0 = w^2 + 12w + 36 - 3 + 2w0=w2+14w+330 = w^2 + 14w + 33
  3. Factor the quadratic: Factor the quadratic equation. \newline(w+11)(w+3)=0(w + 11)(w + 3) = 0
  4. Solve for ww: Solve for ww by setting each factor equal to zero.w+11=0w + 11 = 0 or w+3=0w + 3 = 0w=−11w = -11 or w=−3w = -3
  5. Check for extraneous solutions: Check for extraneous solutions by substituting the values of ww back into the original equation.\newlineCheck w=−11w = -11:\newline3−2(−11)=−11+6\sqrt{3 - 2(-11)} = -11 + 6\newline3+22=−5\sqrt{3 + 22} = -5\newline25=−5\sqrt{25} = -5\newline5≠−55 \neq -5 (This is an extraneous solution because the square root of a positive number cannot be negative.)

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