Find the Non-extraneous Solutions of Square Root X 3-4 X-1
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Mathematics
OpenStudy (anonymous):
Find the non-extraneous solutions of the square root of the quantity x plus 9 minus 5 equals quantity x plus 4. x = −2 x = −9 and x = −2 x = 9 x = −9 and x = −8 --------------------------------------------------------------------------------
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OpenStudy (anonymous):
@youngamerican can you help me on this?
OpenStudy (valpey):
Is this the question?\[\sqrt{x+9}-5=x+4\]
OpenStudy (anonymous):
yes
OpenStudy (valpey):
Start with adding 5 to both sides
OpenStudy (anonymous):
ok
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OpenStudy (valpey):
\[\sqrt{x+9}=x+9\]Now we just have to say that as long as the quantity x+9 is non-negative, then the expression will be true if we square both sides.
OpenStudy (anonymous):
So x will equal 9?
OpenStudy (valpey):
So we square both sides:\[\sqrt{x+9}=x+9\]\[x+9=(x+9)*(x+9)=x^2+18x+81\]Now we simplify to a single quadratic:\[0=x^2+17x+72\]
OpenStudy (valpey):
Now you can use the quadratic equation, but you have to make sure that x+9 >=0
OpenStudy (anonymous):
im so lost
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OpenStudy (valpey):
Okay, you are given possible answers; best to plug and check. the obvious solution is the one where 0=0. Try the others.
OpenStudy (valpey):
One way to think of it is where the square root function (y=sqrt(x)) equals the line y=x. This happens when 0=0 and 1=1.
OpenStudy (anonymous):
math really isn't my subject... thanks for helping and all but I really don't understand what your saying
OpenStudy (valpey):
So you are looking for values of x where x+9 =0 or x+9 = 1
OpenStudy (valpey):
First of all, stop telling yourself that math isn't your subject. The more you tell yourself that, the more you start to believe it. I'm happy to back up though. Did you follow me up through:? \[\sqrt{x+9}=x+9\]
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OpenStudy (anonymous):
yes I got how you added 5 to each side and everything but after that you lost me
OpenStudy (valpey):
Okay, well let's just say that y = x+9 and then (we can just solve for y now; we will solve for x later) we can simplify the equation to \[\sqrt{y}=y\]does that follow?
OpenStudy (valpey):
I'm hoping you are following how I'm temporarily replacing x+9 with a new variable, y. So \[\sqrt{y}=y\]Is basically the same as saying \[y=y^2\] Can you think of any numbers which are equal to their own squares?
OpenStudy (anonymous):
9?
OpenStudy (anonymous):
10?
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OpenStudy (valpey):
10 squared is 100 and 9 squared is 81. In fact any number bigger than 1 will become even larger when multiplied by itself.
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Find the Non-extraneous Solutions of Square Root X 3-4 X-1
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