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

Feb 18, 2017

$\forall k \in \mathbb{R} , f \left(x\right) = 2 {x}^{2} - k x + \left(k - 2\right) = 0$ can not have roots both negative.

#### Explanation:

Let $\alpha < 0 \mathmr{and} \beta < 0$ be the roots of the given quadratic

$f \left(x\right) = 2 {x}^{2} - k x + \left(k - 2\right) = 0 , \left(k \in \mathbb{R}\right) .$

Now, we know that, if $\alpha ' \mathmr{and} \beta '$ are the roots of the quadr.

eqn. $a {x}^{2} + b x + c = 0 ,$ then, $\alpha ' + \beta ' = - \frac{b}{a} , \alpha ' \beta ' = \frac{c}{a} .$

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

$\alpha + \beta = \frac{k}{2.} \ldots . . \left(\ast 1\right) , \mathmr{and} , \alpha \beta = \frac{k - 2}{2.} \ldots . . \left(\ast 2\right) .$

Knowing that, both, $\alpha , \mathmr{and} , \beta < 0 ,$ we must have,

$\alpha + \beta < 0. \ldots \ldots \ldots \left(\ast 1 '\right) , \mathmr{and} , \alpha \beta > 0. \ldots \ldots \ldots \left(\ast 2 '\right)$.

Combining $\left(\ast 1\right) , \left(\ast 1 '\right) , \left(\ast 2\right) \mathmr{and} \left(\ast 2 '\right) ,$ we have,

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

$k < 0 , \mathmr{and} , k > 2 , \text{ a contradiction.}$

We conclude that, $\forall k \in \mathbb{R} , f \left(x\right) = 2 {x}^{2} - k x + \left(k - 2\right) = 0$

can not have roots both negative.

Enjoy Maths.!

Feb 18, 2017

Both roots cannot be negative.
The solutions of this equation is $S = \left\{\frac{k}{2} - 1 , 1\right\}$

#### Explanation:

We need

${a}^{2} - 2 a b + {b}^{2} = {\left(a - b\right)}^{2}$

$2 {x}^{2} - k x + \left(k - 2\right) = 0$

We compare this equation to

$a {x}^{2} + b x + c = 0$

Let's calculate the discriminant

$\Delta = {b}^{2} - 4 a c$

$= {\left(- k\right)}^{2} - 4 \cdot 2 \cdot \left(k - 2\right)$

$= {k}^{2} - 8 k + 16$

$= {\left(k - 4\right)}^{2}$

So,

the roots of the equation are

$x = \frac{- b \pm \sqrt{\Delta}}{2 a}$

$x = \frac{k \pm \sqrt{{\left(k - 4\right)}^{2}}}{2 \cdot 2}$

$= \frac{k \pm \left(k - 4\right)}{4}$

${x}_{1} = \frac{k + k - 4}{4} = \frac{2 k - 4}{4} = \frac{k}{2} - 1$

${x}_{2} = \frac{k - k + 4}{4} = \frac{4}{4} = 1$

Feb 18, 2017

See below.

#### Explanation:

Note that $\left(x - {r}_{1}\right) \left(x - {r}_{2}\right) = {x}^{2} + \left({r}_{1} + {r}_{2}\right) x + {r}_{1} {r}_{2}$

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

Analyzing ${x}^{2} - \frac{k}{2} x + \frac{k - 2}{2}$ the condition states that

$- \frac{k}{2} > 0$ and $\frac{k - 2}{2} > 0$ which is contradictory because

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

Feb 18, 2017

No

#### Explanation:

More briefly, given:

$2 {x}^{2} - k x + \left(k - 2\right) = 0$

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

$2 - k + k - 2 = 0$

Hence $x = 1$ is a zero.