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Mathematics 14 Online
OpenStudy (anonymous):

Debbie and Trisha are looking at the equation the square root of the quantity of 4 times x minus 6 equals square root of x . Debbie says that the solution is extraneous. Trisha says that the solution is non-extraneous. Is Debbie correct? Is Trisha correct? Are they both correct? Justify your response by solving this equation, explaining each step with complete sentences.

OpenStudy (anonymous):

Someone please help! I will reward medal and fan!!

OpenStudy (anonymous):

@grex10

OpenStudy (anonymous):

Do you get this? @leon549

OpenStudy (anonymous):

yeah working on it

OpenStudy (anonymous):

extraneous A solution to an equation that SEEMS to be right, but when we check it (by substituting it into the original equation) turns out NOT to be right.

OpenStudy (anonymous):

So then who is correct?

OpenStudy (anonymous):

@lilcj2001

OpenStudy (anonymous):

OpenStudy (anonymous):

i will go with debbie

OpenStudy (anonymous):

Why?

OpenStudy (anonymous):

the equation will become 4x -6= sqrt x

OpenStudy (anonymous):

square both sides to get a general equation

OpenStudy (xapproachesinfinity):

hey there, let's work it out and find what is happening

OpenStudy (xapproachesinfinity):

\[\sqrt{4x-6}=\sqrt{x}\] this is our equation yes

OpenStudy (xapproachesinfinity):

as leon said we take square of both sides to solve this

OpenStudy (xapproachesinfinity):

\[(\sqrt{4x-6})^2=(\sqrt{x})^2 \Longrightarrow 4x-6=x \] \[4x-x=-6\] can you solve this one?

OpenStudy (xapproachesinfinity):

should be 4x-x=6 not -6 sorry

OpenStudy (xapproachesinfinity):

\[4x-x=6 \]

OpenStudy (xapproachesinfinity):

so can you solve it or need help with that one?

OpenStudy (xapproachesinfinity):

well @LexiLuvv2431 what do say?

OpenStudy (anonymous):

Heyy thank you. Umm I kinda think I get it but at the same time I dont know where to start.

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