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Mathematics 14 Online
OpenStudy (anonymous):

State if it have solution or no solution 4y^2+169=0

OpenStudy (dmezzullo):

Solution in \(\mathbb R\)?

OpenStudy (dmezzullo):

i mean reals?

OpenStudy (anonymous):

Yes

hartnn (hartnn):

Compare your quadratic equation with \(ay^2+by+c=0\) find \[a=...?\\b=...?\\c=...?\] if \(b^2-4ac \ge 0\), then it has a solution, if \(b^2-4ac <0\) then no solution

hartnn (hartnn):

because square root of negative numbers are not defined in reals...

OpenStudy (anonymous):

It has no solition?

OpenStudy (anonymous):

Solution*

hartnn (hartnn):

that would be correct. just curious....what did you get for a,b,c ??

OpenStudy (anonymous):

a=4 b=0 c=169?

hartnn (hartnn):

thats correct! good work then :)

OpenStudy (anonymous):

Thanks!

hartnn (hartnn):

welcome ^_^

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