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

4 Answers
Feb 18, 2017

Answer:

#AA k in RR, f(x)=2x^2-kx+(k-2)=0# can not have roots both negative.

Explanation:

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.!

Feb 18, 2017

Answer:

Both roots cannot be negative.
The solutions of this equation is #S={k/2-1, 1}#

Explanation:

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#

Feb 18, 2017

Answer:

See below.

Explanation:

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.

Feb 18, 2017

Answer:

No

Explanation:

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.