explain why a quadratic equation can't have one imaginary number

[kristiekayc]

explain why a quadratic equation can't have one imaginary number

 

[pka]

explain why a quadratic equation can't have one imaginary number

That is a false statement.
The quadratic equation $\displaystyle x^2-(1+i)x+i=0$ has two roots, $\displaystyle 1~\&~i$.

The statement should should read a quadratic equation with real coefficients can't have only one imaginary root.

The reason being in $\displaystyle x^2+ax+c=0$ because $\displaystyle -a$ is sum of the roots and $\displaystyle c$ is product of the roots.
But $\displaystyle a~\&~c$ are both real numbers, that is impossible if only one of the roots were imaginary.

 

[HallsofIvy]

Another way of looking at it: if quadratic equation, $\displaystyle ax^2+ bx+ c= 0$ has roots, $\displaystyle x_0$ and $\displaystyle x_1$, then we must have $\displaystyle a(x- x_0)(x- x-1)= ax^3- a(x_0+ x_1)x+ ax_0x_1$ so that $\displaystyle a(x_0+ x_1)= b$ and $\displaystyle ax_0x_1= c$. If all coefficents, a, b, and c are real, then $\displaystyle x_0+ x_1= b/c$ and $\displaystyle x_0x_1= c/a$ must be real.

From that it follows that if there is one non-real (strictly speaking not just "imaginary") root, then the other must not only be non-real but must be the complex conjugate of the other.

Yet another way to see it: if $\displaystyle ax^2+ bx+ c= 0$, then, by the quadratic formula, $\displaystyle x= -\frac{b}{2a}\pm\frac{\sqrt{b^2- 4ac}}{2a}$. In order that there be any non-real root, $\displaystyle b^2- 4ac$ must be negative. And in that case, the "$\displaystyle \pm$ must give two complex conjugate complex numbers.

 

[JeffM]

explain why a quadratic equation can't have one imaginary number

That is a VERY good question.

Look at the quadratic formula: $\displaystyle ax^2 + bx + c \implies x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \dfrac{-b}{2a} \pm \sqrt{\dfrac{b^2 - 4ac}{4a^2}}.$ With me so far?

$\displaystyle But \dfrac{-b}{2a}\ is\ just\ one\ single\ number;\ call\ it\ d.$

$\displaystyle And \dfrac{b^2 - 4ac}{4a^2}\ is\ also\ just\ one\ single\ number;\ call\ it\ e^2.$

$\displaystyle So\ x = d \pm \sqrt{e^2} \implies x = d + e\ or\ x = d - e.$

$\displaystyle e^2 = 0 \implies \sqrt{0} = 0 \implies x = d + 0 = d\ or\ x = d - 0 = d\ so\ x = d.\ ONE\ answer\ if\ e^2 = 0.$

$\displaystyle e^2 \ne 0 \implies d + e \ne d - e\ so\ TWO\ different\ possible\ answers\ if\ e^2 \ne 0.$ Right?

$\displaystyle e^2 < 0 \implies e\ is\ "imaginary"\ and \ne 0 \implies d + e \ne d - e\ \implies\ two\ DIFFERENT\ imaginary\ numbers\ are\ possible\ answers.$

Plus and minus lead to two different possible roots of a quadratic unless e² = 0. If e² > 0, both roots are real. If e² < 0, both roots are complex (meaning that they include an "imaginary" part). Make sense now?

PS Looking at PKA's and Hall's posts, I see I should have made clear that I was talking about a quadratic with real coefficients. I tend to make the dangerous assumption that we are dealing with real numbers unless it is disclosed otherwise.

 

[lookagain]

Another way of looking at it: if quadratic equation,

$\displaystyle ax^2+ bx+ c= 0$ has roots,

$\displaystyle x_0$ and $\displaystyle x_1$, then we must have

> > > $\displaystyle a(x- x_0)(x- x-1)= ax^3- a(x_0+ x_1)x+ ax_0x_1$ < < <


so that $\displaystyle a(x_0+ x_1)= b$ and $\displaystyle ax_0x_1= c$.

Typos are contained in the highlighted line above.


Here is an amendment:


"$\displaystyle a(x - x_0)(x - x_1) \ = \ ax^2 - a(x_0 + x_1)x + ax_0x_1$"


_______________________________________________________________



JeffM typed:

"Look at the quadratic formula: > > > $\displaystyle ax^2 + bx + c \implies x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \dfrac{-b}{2a} \pm \sqrt{\dfrac{b^2 - 4ac}{4a^2}}.$ " < < <


Again (as there is a history), implications are from one equation to one equation.
Also, they are not used from an expression to an equation.


Here is a possible amendment:


"Look at the quadratic formula:


$\displaystyle ax^2 + bx + c \ = \ 0 \implies$


$\displaystyle x \ = \ \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} \implies$


$\displaystyle x \ = \ \dfrac{-b}{2a} \pm \sqrt{\dfrac{b^2 - 4ac}{4a^2}}.$"