Question

Simplest way to solve this ?

Answer 1

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`

Answer 2

`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`

Answer 3

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

`+-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`.