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OpenStudy (anonymous):
solve
sqrt9x+55=x+5
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OpenStudy (amistre64):
if you could graph them, the points where they intersect are the solutions
OpenStudy (amistre64):
otherwise, you have to square both sides and work it from that angle and dbl chk the results for extranuous results
jhonyy9 (jhonyy9):
sqrt9x or sqrt(9x+55) ???
OpenStudy (anonymous):
it sqrt 9x+55
OpenStudy (anonymous):
so then it would be 0 right that is what i got
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jhonyy9 (jhonyy9):
after squared both sides you get 9x+55=x2 +10x +25
x2 +x -30=0
x_1,_2=(1+/- sqrt(1+120))/2 =(1+/- sqrt121)/2 = -5 and 6
OpenStudy (amistre64):
\[\sqrt{55}\ne5\]
OpenStudy (anonymous):
so then where did i go wrong if i got 0 instead of sqrt 55 =5
OpenStudy (anonymous):
man i messed that up let me try again
\[9x+55=x^2+10x+25\]
\[x^2+x-30=0\]
\[(x-5)(x+6)=0\]
\[x=5,x=-6\] now we check
OpenStudy (anonymous):
replace x by 5 to get
\[\sqrt{9\times 5+55}=5+5\]
\[\sqrt{100}=10\]
\[10=10\] that one checks
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OpenStudy (anonymous):
\[\sqrt{9\times (-6)+55}=-6+5\]
\[\sqrt{1}=-1\] nope
OpenStudy (anonymous):
so answer is
\[x=5\]
OpenStudy (anonymous):
thank you
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