If both of the zeros of #ax^2+bx+c# are positive or both negative then what can you say about the signs of #a, b# and #c# ?

2 Answers
Aug 6, 2017

Answer:

See explanation...

Explanation:

Given a quadratic polynomial:

#ax^2+bx+c#

  • If the zeros are both positive then #a# and #c# have the same sign and #b# has the opposite sign.

  • If the zeros are both negative then #a#, #b# and #c# have the same sign.

To see these, suppose the zeros are #x=alpha# and #x=beta#.

We have:

#ax^2+bx+c = a(x-alpha)(x-beta)#

#color(white)(ax^2+bx+c) = a(x^2-(alpha+beta)x+alphabeta)#

#color(white)(ax^2+bx+c) = ax^2-(alpha+beta)ax+alphabetaa#

Matching the coefficients, we find:

#{ (b = -(alpha+beta)a), (c = alphabetaa) :}#

If #alpha > 0# and #beta > 0# then #-(alpha+beta) < 0# and #alphabeta > 0#. Hence #b# has the opposite sign to #a# and #c# has the same sign as #a#.

If #alpha < 0# and #beta < 0# then #-(alpha+beta) > 0# and #alphabeta > 0#. Hence #a#, #b# and #c# all have the same sign.

#color(white)()#
Note

The converses of these two results are not generally true.

There are quadratic equations #ax^2+bx+c# with #a, c# having the same sign and #b# having the same or opposite sign which have no real zeros.

For example:

#x^2+x+1#

has no real zeros.

It still has two zeros, but they are so called "imaginary" a.k.a. non-real complex numbers. They do not lie on the real number line, but somewhere else in the complex plane. The real number line forms the #x# axis of the complex plane, the #y# axis being called the imaginary axis.

All complex numbers can be expressed in the form #x+yi#, where #x, y# are real numbers and #i# is the imaginary unit, satisfying:

#i^2 = -1#

For example, the zeros of #x^2+x+1# are:

#-1/2+sqrt(3)/2i" "# and #" "-1/2-sqrt(3)/2i#

It is unfortunate that the term imaginary is used of such numbers - coined when such numbers were poorly understood. They are just as real as Real numbers and better behaved in some respects.

Aug 10, 2017

Answer:

Some more explanation...

Explanation:

Attempting to make this a little easier to follow, let us start with the simpler case where the leading coefficient is #1#

#x^2+px+q#

If #alpha# and #beta# are the two zeros, then both #(x-alpha)# and #(x-beta)# must be factors.

Using FOIL to help multiply, we find:

#(x-alpha)(x-beta) = overbrace((x * x))^"First"+overbrace(x * (-beta))^"Outside"+overbrace((-alpha) * x)^"Inside"+overbrace((-alpha) *(-beta))^"Last"#

#color(white)((x-alpha)(x-beta)) = x^2-betax-alphax+alphabeta#

#color(white)((x-alpha)(x-beta)) = x^2-(alpha+beta)x+alphabeta#

Comparing this with the expression #x^2+px+q# we find:

  • If #alpha > 0# and #beta > 0# then #-(alpha+beta) < 0#, so #p < 0#.
    Also #alphabeta > 0#, so #q > 0#.#color(white)(0/0)#
    In summary, the pattern of the signs of the coefficients of #x^2+px+q# is #+ - +#.#color(white)(0/0)#
    For example #x^2-3x+2 = (x-1)(x-2)# has zeros #1# and #2#.#color(white)(0/0)#

  • If #alpha < 0# and #beta < 0# then #-(alpha+beta) > 0#, so #p > 0#.#color(white)(0/0)#
    Also #alphabeta > 0#, so #q > 0#.#color(white)(0/0)#
    In summary, the pattern of the signs of the coefficients of #x^2+px+q# is #+ + +#.#color(white)(0/0)#
    For example #x^2+3x+2 = (x+1)(x+2)# has zeros #-1# and #-2#.

The more general case of #ax^2+bx+c# is similar, but we have to deal with that initial coefficient #a#, which can be positive or negative.

We have:

#ax^2+bx+c = a(x^2+b/ax+c/a) = a(x-alpha)(x-beta)#

  • If #alpha > 0# and #beta > 0# then the coefficients #1#, #b/a# and #c/a# have signs in the pattern #+ - +#.#color(white)(0/0)#
    Hence if #a > 0# then the pattern of the signs of the coefficients of #ax^2+bx+c# is also #+ - +#.#color(white)(0/0)#
    If instead #a < 0# then the pattern of the signed of the coefficients of #ax^2+bx+c# is #- + -#.#color(white)(0/0)#

  • If #alpha < 0# and #beta < 0# then the coefficients #1#, #b/a# and #c/a# have signs in the pattern #+ + +#.#color(white)(0/0)#
    Hence if #a > 0# then the pattern of the signs of the coefficients of #ax^2+bx+c# is also #+ + +#.#color(white)(0/0)#
    If instead #a < 0# then the pattern of the signs of the coefficients of #ax^2+bx+c# is #- - -#.