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mestny [16]
2 years ago
14

Avery solves the equation below by first squaring both sides of the equation.

Mathematics
2 answers:
muminat2 years ago
5 0

Answer:

z=\dfrac{7}{3}

Step-by-step explanation:

Given equation:

\sqrt{z^2+8}=1-2z

Square both sides:

\implies (\sqrt{z^2+8})^2=(1-2z)^2

\implies z^2+8=1-4z+4z^2

Subtract z^2 from both sides:

\implies 8=1-4z+3z^2

Subtract 8 from both sides:

\implies 0=-7-4z+3z^2

\implies 3z^2-4z-7=0

Rewrite the middle term:

\implies 3z^2+3z-7z-7=0

Factor the first two terms and the last two terms separately:

\implies 3z(z+1)-7(z+1)=0

Factor out the common term (z+1):

\implies (3z-7)(z+1)=0

Therefore:

\implies (z+1)=0 \implies z=-1

\implies (3z-7)=0 \implies z=\dfrac{7}{3}

To find the extraneous solution (the root that is <u>not</u> a root of the original equation), enter the two found values of z into the original equation:

\begin{aligned}z=-1\implies \sqrt{(-1)^2+8}&=1-2(-1)\\\implies 3&=3\implies \textsf{true}\\\end{aligned}

\begin{aligned}z=\dfrac{7}{3} \implies \sqrt{\left(\frac{7}{3}\right)^2+8}&=1-2\left(\frac{7}{3}\right)\\\implies \dfrac{11}{3} &=-\dfrac{11}{3}\implies \textsf{false}\end{aligned}

\textsf{As}\: \dfrac{11}{3} \neq -\dfrac{11}{3}\:\textsf{then}\: z=\dfrac{7}{3}\:\textsf{is the extraneous solution}

8090 [49]2 years ago
4 0

Answer:

z = 7/3 is extraneous.

Step-by-step explanation:

√(z^2+8) = 1 - 2z

z^2 + 8 = (1 - 2z)^2

z^2 + 8 = 1 -4z + 4z^2

3z^2 - 4z - 7 = 0

3z^2 + 3z - 7z - 7 = 0

3z(z + 1) - 7(z + 1) = 0

(3z - 7)(z + 1) = 0

3z = 7, z = -1

z = 7/3, -1.

One of these might be extraneous.

Checking:

√(z^2+8) = 1 - 2z, if z = -1:

√(1 + 8) = 3, -3

1 - 2(-1) = 3.    So its not z = -1

if z = 7/3

√((7/3)^2 + 8) = 13.44

1 - 2(7/3) = -3.66

So its z = 7/3

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