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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 :

MrRoyalidn

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

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