Simplest way to solve this ?

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Simplest way to solve this ?

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Answer 1 · 2018-01-28T02:52:48

I got $p = \pm 6$ $p=+-6$ as the possible values. Details follow...

Explanation:

According to the quadratic formula, the roots are:

$x = \frac{- p \pm \sqrt{p^{2} - 4 (2) (1)}}{2 (2)} = \frac{- p \pm \sqrt{p^{2} - 8}}{4}$ $x=(-p +- sqrt(p^2-4(2)(1)))/(2(2))=(-p+-sqrt(p^2-8))/4$

These roots are labelled $α$ $alpha$ and $β$ $beta$ .

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

${(\frac{- p + \sqrt{p^{2} - 8}}{4})}^{2} + {(\frac{- p - \sqrt{p^{2} - 8}}{4})}^{2} = 8$ $((-p+sqrt(p^2-8))/4)^2 + ((-p-sqrt(p^2-8))/4)^2 = 8$

Simpler to multiply both sides by $4^{2}$ $4^2$ to remove the denominators:

${(- p + \sqrt{p^{2} - 8})}^{2} + {(- p - \sqrt{p^{2} - 8})}^{2} = 128$ $(-p+sqrt(p^2-8))^2 + (-p-sqrt(p^2-8))^2 = 128$

Using FOIL to expand the binomials:

$p^{2} - 2 p \sqrt{p^{2} - 8} + (p^{2} - 8) + p^{2} + 2 p \sqrt{p^{2} - 8} + (p^{2} - 8) = 128$ $p^2 - 2psqrt(p^2-8)+(p^2-8) + p^2 + 2psqrt(p^2-8)+(p^2-8)=128$

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

$2 p^{2} + 2 (p^{2} - 8) = 128$ $2p^2 + 2(p^2-8)=128$

or

$p^{2} + p^{2} - 8 = 64$ $p^2+p^2-8=64$

$2 p^{2} = 72$ $2p^2 = 72$

$p^{2} = 36$ $p^2 = 36$

until, at last $p = \pm 6$ $p=+-6$

Answer 2 · 2018-01-28T08:12:06

$p = \pm 6$ $p=+-6$

Explanation:

$2 x^{2} + p x + 1 = 2 (x - α) (x - β)$ $2x^2+px+1 = 2(x-alpha)(x-beta)$

$2 x^{2} + p x + 1 = 2 x^{2} - 2 (α + β) x + 2 α β$ $color(white)(2x^2+px+1) = 2x^2-2(alpha+beta)x+2alphabeta$

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:

$p = +-6$

Answer 3 · 2018-01-28T08:13:48

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, $p = +-6$

Answer 4 · 2018-01-28T11:09:50

$+-6$ .

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,

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

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

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

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