Is the quadratic formula always right?

[Jason76]

It is always produces the answers to a quadratic, unless it isn't factorable.

 

[DrPhil]

It is always produces the answers to a quadratic, unless it isn't factorable.

Yes. If answers exist, the quadratic formula will find them whether factorable or not. Look at the "discriminant," D = b^2 - 4ac.

If D is positive, there are two roots of the quadratic
....and if D is also a perfect square, the quadratic is factorable.
If D=0, the two roots are equal - that is, the root is double
....and the quadratic is a perfect square.
If D < 0, the two roots are a conjugate pair of complex numbers -
....no real roots.

 

[JeffM]

It is always produces the answers to a quadratic, unless it isn't factorable.

It produces the correct roots for any quadratic, whether the quadratic is factorable or not and whether or not the roots are real or not.

If you know how the formula is derived, the above statements become obvious.

\(\displaystyle Let\ a,\ b,\ and\ c \in \mathbb C\,\ a \ne 0,\ and\ ax^2 + bx + c = 0.\)

\(\displaystyle So\ \dfrac{ax^2}{a} + \dfrac{bx}{a} + \dfrac{c}{a} = \dfrac{0}{a}.\) Divide by a.

\(\displaystyle x^2 + \left(\dfrac{2}{2} * \dfrac{bx}{a}\right) + \left(\dfrac{4a}{4a} * \dfrac{c}{a}\right) = 0.\) Nothing weird here.

\(\displaystyle x^2 + 2\left(\dfrac{b}{2a}\right) + \dfrac{4ac}{4a^2} = 0 \implies x^2 + 2\left(\dfrac{b}{2a}\right) = - \dfrac{4ac}{4a^2}.\) Just algebra.

\(\displaystyle x^2 + 2\left(\dfrac{b}{2a}\right) + \dfrac{b^2}{4a^2}= \dfrac{b^2}{4a^2} - \dfrac{4ac}{4a^2}.\) More algebra.

\(\displaystyle x^2 + 2\left(\dfrac{b}{2a}\right) + \left(\dfrac{b}{2a}\right)^2 = \dfrac{b^2 - 4ac}{4a^2}.\) And more.

\(\displaystyle \left(x + \dfrac{b}{2a}\right)^2 = \dfrac{b^2 - 4ac}{4a^2}.\) And more

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

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

Nothing about factoring or real numbers in the derivation.

 

[Deleted member 4993]

It is always produces the answers to a quadratic, unless it isn't factorable.

Your statement - as posted - does not make sense!

The quadratic formula will provide roots of a quadratic function (or solutions to a quadratic equation) - even if the expression is not factorisable within real (and/or rational) domain.

 

[Jason76]

Your statement - as posted - does not make sense!

The quadratic formula will provide roots of a quadratic function (or solutions to a quadratic equation) - even if the expression is not factorisable within real (and/or rational) domain.

So does the quadratic equation ALWAYS have roots? Roots which can always be found using the quadratic formula, whether the source equation is factorable or not.

 

[Deleted member 4993]

So does the quadratic equation ALWAYS have roots?

Roots which can always be found using the quadratic formula, whether the source equation is factorable or not.

Equations have solutions - functions have roots.

Yes ... like DrPhil and JeffM said - those roots or solutions may be irrational numbers and/or complex numbers.

 

[JeffM]

So does the quadratic equation ALWAYS have roots? Roots which can always be found using the quadratic formula, whether the source equation is factorable or not.

You can think about this two ways:

A quadratic with real coefficients always has two roots, which may not be distinct and which may not be real.

A quadratic with real coefficients has two real roots if the discriminant is positive; one real root if the discriminant is zero, and two complex roots if the discriminant is negative. In all three cases, the quadratic formula gives you the roots.

I prefer the second formulation myself.

 

[HallsofIvy]

In terms of complex numbers, yes. The "Fundamental Theorem of Algebra"- every polynomial equation of degree n can be factored into n linear factors with coefficients in the complex numbers. The quadratic equation, \(\displaystyle x^2+ 1= 0\), for example, has no real number solutions but two complex number solutions, i and -i. And, yes, the quadratic formula will give those solutions. Here, a= 1, b= 0, c= 1 so \(\displaystyle x= \frac{-b\pm\sqrt{b^2- 4ac}}{2a}= \frac{0\pm\sqrt{0- 4(1)(1)}}{2}= \frac{\pm 2i}{2}= \pm i\)