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Consider the equation #sqrt(3t - 14) + t = 6#, are there any extraneous solutions? What is the solution for this equation?

Answer 1

Eva Abramson

Solution: #t=5#
Using standard methods, we may also find the extraneous solution #t=10#

Given #color(white)("XXX")sqrt(3t-14)+t=6#

#color(white)("XXX")sqrt(3t-14)=6-t#

after squaring both sides: #color(white)("XXX")3t-14=36-12t+t^2#

then convert into standard form #color(white)("XXX")t^2-15+50=0#

and factor #color(white)("XXX")(t-5)(t-10)=0#

which implies #{: ("either",(t-5)=0," or ",(t-10)=0), (,rarr t=5,,rarr t=10) :}#

If #t=5# then #color(white)("XXX")sqrt(3t-14)+t=sqrt(3xx5-14)+5=sqrt(1)+5=6# as required #rarr t=5# is not an extraneous solution.

If #t=10# then #color(white)("XXX")sqrt(3t-14)+t=sqrt(3xx10-14)+10=sqrt(16)+10=14!=6# #rarr t=10# is extraneous The problem arises because the square root symbol (#sqrt(color(white)(x")#) is always taken to be the primary (positive) root.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.