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

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

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Answer 1 · 2017-08-06T07:10:16

See explanation...

Explanation:

Given a quadratic polynomial:

$a x^{2} + b x + c$ $ax^2+bx+c$

If the zeros are both positive then $a$ $a$ and $c$ $c$ have the same sign and $b$ $b$ has the opposite sign.
If the zeros are both negative then $a$ $a$ , $b$ $b$ and $c$ $c$ have the same sign.

To see these, suppose the zeros are $x = α$ $x=alpha$ and $x = β$ $x=beta$ .

We have:

$a x^{2} + b x + c = a (x - α) (x - β)$ $ax^2+bx+c = a(x-alpha)(x-beta)$

$a x^{2} + b x + c = a (x^{2} - (α + β) x + α β)$ $color(white)(ax^2+bx+c) = a(x^2-(alpha+beta)x+alphabeta)$

$a x^{2} + b x + c = a x^{2} - (α + β) a x + α β a$ $color(white)(ax^2+bx+c) = ax^2-(alpha+beta)ax+alphabetaa$

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.

Answer 2 · 2017-08-10T18:35:51

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 $- - -$ .

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

2 Answers

Explanation:

Explanation:

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If both of the zeros of ax^2+bx+cax2+bx+cax^2+bx+c are positive or both negative then what can you say about the signs of a, ba,ba, b and ccc ?

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Explanation:

Explanation:

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