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:


\(\displaystyle ax^2 + bx + c \ = \ 0 \implies \)


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


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