Simplest way to solve this ?

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4 Answers
Jan 28, 2018

I got #p=+-6# as the possible values. Details follow...

Explanation:

According to the quadratic formula, the roots are:

#x=(-p +- sqrt(p^2-4(2)(1)))/(2(2))=(-p+-sqrt(p^2-8))/4#

These roots are labelled #alpha# and #beta#.

I will square both, add them and set the sum equal to 8:

#((-p+sqrt(p^2-8))/4)^2 + ((-p-sqrt(p^2-8))/4)^2 = 8#

Simpler to multiply both sides by #4^2# to remove the denominators:

#(-p+sqrt(p^2-8))^2 + (-p-sqrt(p^2-8))^2 = 128#

Using FOIL to expand the binomials:

#p^2 - 2psqrt(p^2-8)+(p^2-8) + p^2 + 2psqrt(p^2-8)+(p^2-8)=128#

The two radicals on the left cancel (fortunately), leaving

#2p^2 + 2(p^2-8)=128#

or

#p^2+p^2-8=64#

#2p^2 = 72#

#p^2 = 36#

until, at last #p=+-6#

Jan 28, 2018

#p=+-6#

Explanation:

#2x^2+px+1 = 2(x-alpha)(x-beta)#

#color(white)(2x^2+px+1) = 2x^2-2(alpha+beta)x+2alphabeta#

So:

#{ (alpha+beta=-p/2), (alphabeta=1/2) :}#

Then:

#8 = alpha^2+beta^2 = (alpha+beta)^2-2alphabeta = p^2/4-1#

So:

#p^2 = 4(8+1) = 36 = 6^2#

So:

#p = +-6#

Jan 28, 2018

See below

Explanation:

#2x^2 + px + 1 = 0#

We know,

#alpha + beta = -b/a#

#alpha + beta = -p/2#

#alpha * beta = c/a#

#alpha * beta = 1/2#

#alpha ^2 + beta ^2 = 8#

#(alpha + beta)^2 - 2alpha*beta = 8#

Using the above values

#(-p/2)^2 - 2*1/2 = 8#

#(-p/2)^2 - 1 = 8#

#(-p/2)^2 = 9#

# -p/2 = +-3#

# -p/2 = 3#

# -p = 6#

# p = -6#

# -p/2 = -3#

# -p = -6#

# p = 6#

So, #p = +-6#

Jan 28, 2018

# +-6#.

Explanation:

#alpha and beta# are the roots of the qudr. eqn. #2x^2+px+1=0#.

#:. alpha+beta=-p/2.................(ast)#.

Also, #alpha and beta# must satisfy the quadr. eqn.

#:. 2alpha^2+palpha+1=0, and, 2beta^2+pbeta+1=0#.

Adding these, we get,

#2(alpha^2+beta^2)+p(alpha+beta)+2=0#.

#(ast) and alpha^2+beta^2=8 rArr 2(8)+p(-p/2)+2=0, i.e.,#

#18=p^2/2, or, p^2=36," giving, "p=+-6#.