Mathematics
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OpenStudy (anonymous):
HELP ME
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OpenStudy (anonymous):
\[\sqrt{5X+1}-\sqrt{X}=5?\]
OpenStudy (anonymous):
NEED SOLUTION solve for x
OpenStudy (anonymous):
\[\sqrt{5x+1}=5+\sqrt{X}\]
OpenStudy (hba):
1) add sqrt(x) on both sides
2) Squaring both sides
OpenStudy (anonymous):
ready
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OpenStudy (hba):
What ?
OpenStudy (unklerhaukus):
ready , set , go!
OpenStudy (anonymous):
ll be 5x+1=25+x
OpenStudy (hba):
No Use (a+b)^2=a^2+2ab+b^2
OpenStudy (anonymous):
okay ll be 5x+1=(5+\[\sqrt{x}\])^2
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OpenStudy (hba):
Now use the formula ?
OpenStudy (anonymous):
\[5x+1=(5+\sqrt{x})^2\]
OpenStudy (anonymous):
im i r8?@hba
OpenStudy (hba):
Yes but now use
(a+b)^2=a^2+2ab+b^2
what will
[5+sqrt(x)]^2 be ?
OpenStudy (anonymous):
ll be \[(5+\sqrt{x})^2=25+10\sqrt{x}+x\]
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OpenStudy (hba):
Put it in your question now.
OpenStudy (anonymous):
ll be\[5x+1=25+10\sqrt{x}+x\]
OpenStudy (anonymous):
then bro?
OpenStudy (anonymous):
\[5x-x=25-1+10\sqrt{x}\]
OpenStudy (anonymous):
\[4x=24+10\sqrt{x}\]
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OpenStudy (anonymous):
then?
OpenStudy (hba):
4x-24=10sqrt(x)
2(2x-12)=10sqrt(x)
2x-12=10sqrt(x)/2
2x-12=5sqrt(x)
Now Use Squaring both sides again.
OpenStudy (anonymous):
\[(2x-12)^2=5\]
OpenStudy (anonymous):
\[4x^2+48x+144=5\]
OpenStudy (anonymous):
\[4x^2+48x+139=0\]
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OpenStudy (anonymous):
can we use\[x=(-b \pm \sqrt{b^2-4ac})/2a?\]
OpenStudy (hba):
Now use Quad formula
or
Breaking the middle term
or
Completing the square method
OpenStudy (hba):
Yes you can use Quad formula :)
OpenStudy (hba):
No don't be lazy :/
OpenStudy (hba):
@mustry
Start working.
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OpenStudy (anonymous):
\[x=(-48\pm \sqrt{2304-2224})/8\]
OpenStudy (anonymous):
\[x=(-48\pm \sqrt{80})/8\]
OpenStudy (hba):
I guess you can simplify it further
OpenStudy (hba):
@mustry
I guess you are done :P
OpenStudy (anonymous):
\[x=(-48\pm8.9)/8\]
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OpenStudy (anonymous):
the answer have been given to be either 0 or 16
OpenStudy (anonymous):
when u put 16 u ll get correct but how can we get it please
OpenStudy (hba):
I can only instruct you :/
OpenStudy (anonymous):
okay