# 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 ?

Aug 6, 2017

See explanation...

#### Explanation:

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

• 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:

$a {x}^{2} + b x + c = a \left(x - \alpha\right) \left(x - \beta\right)$

$\textcolor{w h i t e}{a {x}^{2} + b x + c} = a \left({x}^{2} - \left(\alpha + \beta\right) x + \alpha \beta\right)$

$\textcolor{w h i t e}{a {x}^{2} + b x + c} = a {x}^{2} - \left(\alpha + \beta\right) a x + \alpha \beta a$

Matching the coefficients, we find:

$\left\{\begin{matrix}b = - \left(\alpha + \beta\right) a \\ c = \alpha \beta a\end{matrix}\right.$

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

If $\alpha < 0$ and $\beta < 0$ then $- \left(\alpha + \beta\right) > 0$ and $\alpha \beta > 0$. Hence $a$, $b$ and $c$ all have the same sign.

$\textcolor{w h i t e}{}$
Note

The converses of these two results are not generally true.

There are quadratic equations $a {x}^{2} + b x + c$ 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 $x + y i$, 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:

$- \frac{1}{2} + \frac{\sqrt{3}}{2} i \text{ }$ and $\text{ } - \frac{1}{2} - \frac{\sqrt{3}}{2} i$

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

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 $1$

${x}^{2} + p x + q$

If $\alpha$ and $\beta$ are the two zeros, then both $\left(x - \alpha\right)$ and $\left(x - \beta\right)$ must be factors.

Using FOIL to help multiply, we find:

$\left(x - \alpha\right) \left(x - \beta\right) = {\overbrace{\left(x \cdot x\right)}}^{\text{First"+overbrace(x * (-beta))^"Outside"+overbrace((-alpha) * x)^"Inside"+overbrace((-alpha) *(-beta))^"Last}}$

$\textcolor{w h i t e}{\left(x - \alpha\right) \left(x - \beta\right)} = {x}^{2} - \beta x - \alpha x + \alpha \beta$

$\textcolor{w h i t e}{\left(x - \alpha\right) \left(x - \beta\right)} = {x}^{2} - \left(\alpha + \beta\right) x + \alpha \beta$

Comparing this with the expression ${x}^{2} + p x + q$ we find:

• If $\alpha > 0$ and $\beta > 0$ then $- \left(\alpha + \beta\right) < 0$, so $p < 0$.
Also $\alpha \beta > 0$, so $q > 0$.$\textcolor{w h i t e}{\frac{0}{0}}$
In summary, the pattern of the signs of the coefficients of ${x}^{2} + p x + q$ is $+ - +$.$\textcolor{w h i t e}{\frac{0}{0}}$
For example ${x}^{2} - 3 x + 2 = \left(x - 1\right) \left(x - 2\right)$ has zeros $1$ and $2$.$\textcolor{w h i t e}{\frac{0}{0}}$

• If $\alpha < 0$ and $\beta < 0$ then $- \left(\alpha + \beta\right) > 0$, so $p > 0$.$\textcolor{w h i t e}{\frac{0}{0}}$
Also $\alpha \beta > 0$, so $q > 0$.$\textcolor{w h i t e}{\frac{0}{0}}$
In summary, the pattern of the signs of the coefficients of ${x}^{2} + p x + q$ is $+ + +$.$\textcolor{w h i t e}{\frac{0}{0}}$
For example ${x}^{2} + 3 x + 2 = \left(x + 1\right) \left(x + 2\right)$ has zeros $- 1$ and $- 2$.

The more general case of $a {x}^{2} + b x + c$ is similar, but we have to deal with that initial coefficient $a$, which can be positive or negative.

We have:

$a {x}^{2} + b x + c = a \left({x}^{2} + \frac{b}{a} x + \frac{c}{a}\right) = a \left(x - \alpha\right) \left(x - \beta\right)$

• If $\alpha > 0$ and $\beta > 0$ then the coefficients $1$, $\frac{b}{a}$ and $\frac{c}{a}$ have signs in the pattern $+ - +$.$\textcolor{w h i t e}{\frac{0}{0}}$
Hence if $a > 0$ then the pattern of the signs of the coefficients of $a {x}^{2} + b x + c$ is also $+ - +$.$\textcolor{w h i t e}{\frac{0}{0}}$
If instead $a < 0$ then the pattern of the signed of the coefficients of $a {x}^{2} + b x + c$ is $- + -$.$\textcolor{w h i t e}{\frac{0}{0}}$

• If $\alpha < 0$ and $\beta < 0$ then the coefficients $1$, $\frac{b}{a}$ and $\frac{c}{a}$ have signs in the pattern $+ + +$.$\textcolor{w h i t e}{\frac{0}{0}}$
Hence if $a > 0$ then the pattern of the signs of the coefficients of $a {x}^{2} + b x + c$ is also $+ + +$.$\textcolor{w h i t e}{\frac{0}{0}}$
If instead $a < 0$ then the pattern of the signs of the coefficients of $a {x}^{2} + b x + c$ is $- - -$.