How do you solve #4p ^ { 2} + 5p + 2= 0#?

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Answer 1 · 2017-12-26T13:32:16

#=> p = (-5 pm isqrt(7) )/8 #

Your probably used to solving this with #x's# but we can do the sane for #p's#

#p = (-b pm sqrt(b^2 -4ac))/(2a) #

We know #a = 4, b=5 ,c= 2 #

Applying the formula:

#p = (-5 + sqrt( 5^2 - (4*4*2 ) ) )/(2*4) #

#=> p = (-5 pm sqrt(-7) )/(8) #

We know #i = sqrt(-1) #

#=> p = (-5 pm isqrt(7) )/8 #

We can see that there are no real roots by graphing:

enter image source here

Hence only complex solutions!