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Answer 1 · 2015-09-05T03:41:45

Please see the explanation section.

#2x^2+bx+c = 0# are #p# and #2# are zeros.

The Factor Theorem tells us that

#2x^2+bx+c=a(x-p)(x-2)#

Furthermore,

#a(x-p)(x-2) = (ax-ap)(x-2)#

# = ax^2-2ax-apx+2ap#

Now, we have
#ax^2+(-2a-ap)x+2ap = 2x^2+bx+c#

So #a = 2# and

#b = -2a-ap = -4-2p#

Multiply by #(p-2)/(p-2)# to get

#b = -4-2p = (-4-2p )/1 * (p-2)/(p-2)#

# =(8-2p^2)/(p-2)#

Answer 2 · 2015-09-05T08:55:48

Answer is same explanation is different.

For the given quadratic equation sum of the roots is given by the expression #-b/2#, this means that p+2= #-b/2#. Therefore,

b= -2(p+2)= #(-2(p+2)(p-2))/(p-2)# =#(8-2p^2)/(p-2)#