Answer:
[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]
