Question

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

Answer 1

[tex]z=\dfrac{7}{3}[/tex]

Step-by-step explanation:

Given equation:

[tex]\sqrt{z^2+8}=1-2z[/tex]

Square both sides:

[tex]\implies (\sqrt{z^2+8})^2=(1-2z)^2[/tex]

[tex]\implies z^2+8=1-4z+4z^2[/tex]

Subtract [tex]z^2[/tex] from both sides:

[tex]\implies 8=1-4z+3z^2[/tex]

Subtract 8 from both sides:

[tex]\implies 0=-7-4z+3z^2[/tex]

[tex]\implies 3z^2-4z-7=0[/tex]

Rewrite the middle term:

[tex]\implies 3z^2+3z-7z-7=0[/tex]

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

[tex]\implies 3z(z+1)-7(z+1)=0[/tex]

Factor out the common term [tex](z+1)[/tex]:

[tex]\implies (3z-7)(z+1)=0[/tex]

Therefore:

[tex]\implies (z+1)=0 \implies z=-1[/tex]

[tex]\implies (3z-7)=0 \implies z=\dfrac{7}{3}[/tex]

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

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

[tex]\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}[/tex]

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

Answer 2

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