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Mathematics 21 Online
OpenStudy (anonymous):

Find the solutions to the equation (check all that apply) x2+14x+48=0. a.x=-2 b.x=1/2 c.x=2 d.x=-1/2 e.x=3 f.x=3/2

OpenStudy (anonymous):

try plugging each value into the function until you find one that makes it true

OpenStudy (anonymous):

the equations's the same as the last problem, but the choices are different...are you sure about that equation?

OpenStudy (anonymous):

I just need help and the equation is wrog its 4x2+4x+1=0

OpenStudy (anonymous):

a little trick on factoring...some forms afe nicer than others. \[(ax+b)^{2}=a^{2}x^{2}+2abx +b^{2}\] since your first term is a perfect square and your last term is a perfect square, it's a good candidate.

OpenStudy (anonymous):

if it works, then a=2 and b=1. so the question is, does 2ab=4? if so, then your factored form would be \[(ax+b)^{2}\]where a=2 and b=1

OpenStudy (anonymous):

2(2)(1)=4...you're in luck

OpenStudy (anonymous):

so, your factored form is \[(2x+1)^{2}=(2x+1)(2x+1)=0\] so now solve the same as last problem.

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