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goblinko [34]
3 years ago
13

I am confused on numbers 25 and 29, the instructions are at the top.

Mathematics
1 answer:
kvasek [131]3 years ago
3 0
#25 is fairly simple.  Plug in -4 and 3 into the equation, and the extraneous root will be the one that does not work.
\sqrt{12-(-4)} =  \sqrt{16} =  \frac{+}{}4
Extraneous root in this case is positive four since +4≠-4
<span>\sqrt{12-3} = \sqrt{9} = \frac{+}{}3
</span>In this case it's negative 3, since -3≠3

#29 can be turned into a quadratic equation.
x= \sqrt{2x+3}
Square both sides to get
x^{2}=2x+3
Then bring the 2x+3 to the other side, setting the quadratic equal to zero.
x^{2}-2x-3=0
Factor to find that it's equivalent to
(x-3)(x+1)=0
Therefore x is equal to positive 3 and negative 1.  Plug both back into the original equation.  Whichever does not work is the extraneous root, and the answer is the one that does.
<span>x= \sqrt{2x+3}
</span><span>3= \sqrt{2(3)+3}
</span><span>3= \sqrt{9}
</span>Extraneous root would be negative 3.

<span>-1= \sqrt{2(-1)+3}
</span><span>-1= \sqrt{1}
</span>Extraneous root would be positive 1.

Your answers are positive 3 and negative 1.
Extraneous roots are negative 3 and positive 1.
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