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Emily is solving the following equation for 
z.

z-8=sqrt(3z-5)
Her first few steps are given below.

{:[(z-8)^(2)=(sqrt(3z-5))^(2)],[z^(2)-16 z+64=3z-5]:}
Is it necessary for Emily to check her answers for extraneous solutions?
Choose 1 answer:
(A) Yes
(B) No

Emily is solving the following equation for z z .\newlinez8=3z5 z-8=\sqrt{3 z-5} \newlineHer first few steps are given below.\newline(z8)2amp;=(3z5)2z216z+64amp;=3z5 \begin{aligned} (z-8)^{2} & =(\sqrt{3 z-5})^{2} \\ z^{2}-16 z+64 & =3 z-5 \end{aligned} \newlineIs it necessary for Emily to check her answers for extraneous solutions?\newlineChoose 11 answer:\newline(A) Yes\newline(B) No

Full solution

Q. Emily is solving the following equation for z z .\newlinez8=3z5 z-8=\sqrt{3 z-5} \newlineHer first few steps are given below.\newline(z8)2=(3z5)2z216z+64=3z5 \begin{aligned} (z-8)^{2} & =(\sqrt{3 z-5})^{2} \\ z^{2}-16 z+64 & =3 z-5 \end{aligned} \newlineIs it necessary for Emily to check her answers for extraneous solutions?\newlineChoose 11 answer:\newline(A) Yes\newline(B) No
  1. Equation setup: Emily starts with the equation z8=3z5z - 8 = \sqrt{3z - 5}. To solve for zz, she squares both sides of the equation to eliminate the square root.\newline(z8)2=(3z5)2(z - 8)^2 = (\sqrt{3z - 5})^2
  2. Squaring both sides: By squaring both sides, Emily gets:\newlinez216z+64=3z5z^2 - 16z + 64 = 3z - 5\newlineThis is a quadratic equation.
  3. Quadratic equation: Emily needs to bring all terms to one side to set the equation to zero and solve the quadratic equation.\newlinez216z+643z+5=0z^2 - 16z + 64 - 3z + 5 = 0
  4. Combining like terms: Combining like terms, Emily gets: z219z+69=0z^2 - 19z + 69 = 0 This is the simplified form of the quadratic equation.
  5. Solving the quadratic equation: Emily will now solve the quadratic equation for zz using methods such as factoring, completing the square, or the quadratic formula.\newlineHowever, before proceeding, we need to address the original question: Is it necessary for Emily to check her answers for extraneous solutions?
  6. Checking for extraneous solutions: When an equation involves a square root and both sides are squared to eliminate the square root, there is a possibility of introducing extraneous solutions. These are solutions that do not satisfy the original equation.\newlineTherefore, it is necessary for Emily to check her answers against the original equation to ensure they are valid.

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