Given `2x^2-kx+(k-2) = 0`, is there some value of `k` such that both roots are negative?

Let `alpha <0 and beta <0` be the roots of the given quadratic

`f(x)=2x^2-kx+(k-2)=0, (k in RR).`

Now, we know that, if `alpha' and beta'` are the roots of the quadr.

eqn. `ax^2+bx+c=0,` then, `alpha'+beta'=-b/a, alpha'beta'=c/a.`

In our case, then, from this, it follows that,

`alpha+beta=k/2......(ast1) , and, alphabeta=(k-2)/2......(ast2).`

Knowing that, both, `alpha, and, beta <0,` we must have,

`alpha+beta<0..........(ast1'), and, alphabeta>0..........(ast2')`.

Combining `(ast1),(ast1'),(ast2) and (ast2'),` we have,

`k/2<0.......(1), &, (k-2)/2>0........(2),` or, equivalently,

`k<0, and, k>2," a contradiction."`

We conclude that, `AA k in RR, f(x)=2x^2-kx+(k-2)=0`

can not have roots both negative.

Enjoy Maths.!

We need

`a^2-2ab+b^2=(a-b)^2`

`2x^2-kx+(k-2)=0`

We compare this equation to

`ax^2+bx+c=0`

Let's calculate the discriminant

`Delta=b^2-4ac`

`=(-k)^2-4*2*(k-2)`

`=k^2-8k+16`

`=(k-4)^2`

So,

the roots of the equation are

`x=(-b+-sqrtDelta)/(2a)`

`x=(k+-sqrt((k-4)^2))/(2*2)`

`=(k+-(k-4))/(4)`

`x_1=(k+k-4)/4=(2k-4)/4=k/2-1`

`x_2=(k-k+4)/4=4/4=1`

Note that `(x-r_1)(x-r_2) = x^2+(r_1+r_2)x+r_1r_2`

If the trinom has both roots negative then all its coefficients must be positive.

Analyzing `x^2-k/2x+(k-2)/2` the condition states that

`-k/2 > 0` and `(k-2)/2 > 0` which is contradictory because

`k < 0` and `k > 2` are incompatibles. So no both negative roots is possible.

More briefly, given:

`2x^2-kx+(k-2) = 0`

Note that the sum of the coefficients is `0`, that is:

`2-k+k-2 = 0`

Hence `x=1` is a zero.