Answered

Get reliable answers to your questions at Westonci.ca, where our knowledgeable community is always ready to help. Explore our Q&A platform to find reliable answers from a wide range of experts in different fields. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately.

Miku solves the equation below by first squaring both sides of the equation.
\sqrt{z^2+2z-3}=z-3
z
2
+2z−3
​
=z−3square root of, z, squared, plus, 2, z, minus, 3, end square root, equals, z, minus, 3
What extraneous solution does Miku obtain?
z=z=

Sagot :

MrRoyal

Extraneous solutions are untrue solution of an equation

Mike does not obtain any extraneous solutions

How to determine the extraneous solutions

The equation is given as:

[tex]\sqrt{z^2+2z-3}=z-3[/tex]

When squared, it becomes

[tex]z^2+2z-3=z^2-6z + 9[/tex]

Evaluate the like terms

[tex]2z-3=-6z + 9[/tex]

Collect like terns

[tex]6z + 2z = 9 + 3[/tex]

Evaluate the like terms

[tex]8z = 12[/tex]

Divide through by 8

[tex]z = 1.5[/tex]

Substitute 1.5 for z in the given equation

[tex]\sqrt{z^2+2z-3}=z-3[/tex]

[tex]\sqrt{1.5^2+2*1.5-3}=1.5-3[/tex]

Evaluate

[tex]\sqrt{2.25}=-1.5[/tex]

Determine the square root of 2.25

[tex]-1.5=-1.5[/tex]

Hence, Mike does not obtain any extraneous solutions

Read more about extraneous solutions at:

https://brainly.com/question/2959656

We hope our answers were helpful. Return anytime for more information and answers to any other questions you may have. Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. Westonci.ca is committed to providing accurate answers. Come back soon for more trustworthy information.