MHB Quadratic equation with rational roots

AI Thread Summary
The discussion focuses on forming a quadratic equation in the form \( ax^2 + bx + c = 0 \) with integer coefficients \( a, b, c \) in arithmetic progression (AP) that has rational roots. It establishes that for rational roots, the discriminant must be a perfect square, leading to the condition \( (k-3a)^2 - 3(2a)^2 = \Box \). The conversation draws parallels to Pythagorean triples, presenting a parametric solution involving coprime integers \( m \) and \( n \), where \( a, b, c \) are expressed in terms of these integers. A refinement of the definition of a primitive solution is proposed, emphasizing that the common difference \( b-a \) must be positive, ensuring the coefficients remain in AP. Ultimately, the discussion provides a comprehensive parametric form for generating valid quadratic equations with rational roots.
kaliprasad
Gold Member
MHB
Messages
1,333
Reaction score
0
form quadratic equation $ax^2 +bx+c=0$ in parametric form such that a,b,c are integers in AP and it has got rational roots
 
Mathematics news on Phys.org
I worked this out when thinking about http://mathhelpboards.com/challenge-questions-puzzles-28/forming-quadratic-equation-10437.html.

[sp]Let $b = a+k$ and $c = a+2k$. Then the equation is $ax^2 + (a+k)x + (a+2k) = 0$, and the solutions are $x = \frac1{2a}\bigl( -a-k \pm\sqrt{(a+k)^2 - 4a(a+2k)}\bigr).$ The condition for rational roots is that the expression under the square root should be a square, namely $(a+k)^2 - 4a(a+2k) = \Box.$ Write that as $(k-3a)^2 - 3(2a)^2 = \Box$. This shows that we are looking for integer solutions of the equation $x^2 + 3y^2 = z^2.$

That is similar to the Pythagorean triples equation. As in that equation, we look for primitive solutions (those in which $x$, $y$ and $z$ are co-prime). The method is similar to the Pythagorean case, and the parametric formula for the solution is that for any co-prime integers $m$, $n$, with $m$ not a multiple of $3$, we have $x=3n^2 - m^2$, $y = 2mn$ and $z = 3n^2+m^2$. In terms of $a$ and $k$, that gives $a = mn$, $k = m^2+3mn+3n^2$. Finally, if we substitute into the expressions for $b$ and $c$, we see that the primitive solutions are $$a = mn,$$ $$b = m^2 + 4mn + 3n^2,$$ $$c = 2m^2 + 7mn + 6n^2,$$ where (as before) $m$ and $n$ are co-prime and $m$ is not a multiple of $3$.[/sp]
 
kaliprasad said:
form quadratic equation $ax^2 +bx+c=0$ in parametric form such that a,b,c are integers in AP and it has got rational roots
let $a=b-d, c=b+d$
if it has got rational roots
we have :$b^2-4(b+d)(b-d)=b^2-4(b^2-d^2)=4d^2-3b^2=k^2$
here $b,d,k\in Z$
there are many quadratic equatons meet the restriction
in particular if $c=(b+d)=0 \,\,or\,\,(d=-b)$the equation will be:
$2bx^2+bx=0$
and the solutons are $x=0,$ and $x=\dfrac {-1}{2}$
if $b=0 \,\,and\,\,(d\neq 0)$the equation will be:
$-dx^2+d=0$
and the solutons are $x=\pm1$
 
Last edited:
This is to reconcile Albert's solution with mine.

[sp]I defined a primitive solution to be one in which the coefficients $a$, $b$, $c$ have no common factor (other than $\pm1$). In Albert's family of equations $2bx^2 + bx = 0$, the only primitive solutions are those where $b = \pm1.$ In particular, if $b = -1$ then the equation becomes $(-2)x^2 + (-1)x + 0 = 0$, which occurs in my family of primitive solutions by taking $m= -2$ and $n=1$.

That example raises something that I overlooked in my solution. My formula for the common difference $k$ in the AP is $k = m^2 + 3mn + 3n^2$. That expression is positive definite, so my formula only covers equations whose coefficients form an AP in which the common difference is positive. But if $ax^2 + bx + c = 0$ has coefficients forming an AP with negative common difference then by taking its negative you get the equation $(-a)x^2 + (-b)x + (-c) = 0$. That is obviously just a different way of writing the same equation, but this time the coefficients form an AP with positive common difference.

So I need to reformulate the definition of a primitive solution to be one in which $a$, $b$, $c$ have no common factor (other than $\pm1$) and the common difference $b-a$ is positive. Then the formula $$a = mn,$$ $$b = m^2 + 4mn + 3n^2,$$ $$c = 2m^2 + 7mn + 6n^2,$$ where $m$ and $n$ are nonzero co-prime integers, $m$ is not a multiple of $3$, and $n>0$, gives what the OP wanted, namely a parametric form for the primitive solutions. The general solution is then obtained by multiplying all the coefficients in a primitive solution by some nonzero integer $d$.[/sp]
 
my solution

Let the quadratic equation be

$ax^2+bx+c = 0$as the coefficients are in ap so let common difference be y so b-a = y c-b = y

or a = b-y and c = b+ y

for the equation to have rational roots we have

discriminant is a perfect square that is say $n^2$$b^2 – 4 ac = n^2$

or $b^2-4(b-y)(b+y) = n^2$

or $b^2-4(b^2 – y^2) = n^2$

or $4y^2-3b^2= n^2$

dividing by $n^2$ and letting $\frac{b}{n} = t$ and $\frac{y}{n}$ = s we get

$4s^2-3 t^2= 1$ ..1

s and t are real numbers

And from inspection we have a root s = $\frac{1}{2}$, t =0 is a rational root and to find another rational root we draw a straight line from (0,1/2)

We get $s = mt+ \frac{1}{2}$is a general root …(2)

So we put in 1 to get the intersection of the curve given in 1 with this straight line to get

$(2mt+1)^2 - 3t^2 = 1$

Or $4m^2t^2 +4mt+1 – 3t^2 = 1$

Or$ (4m^2-3)t^2 + 4mt = 0$

ignoring the starting solution t = 0 we get

$t= \frac{- 4m}{4m^2-3}$

so $s = mt+ \frac{1}{2}$ we get $s = -\frac{4m^2-8m + 3}{8m^2-6}$

now putting $n = 8m^2-6$ to get rid of denominator we get

b = -8m and $y = -4m^2+3$

so we get

$a = 4m^2 – 8m + 3$
$b = - 8m$
$c = -4m^2 -8m -3$

and taking m as different integers we get different solutions in integers

for example
m = 1 gives (-1,-8, -15) multiply by one we get x^2+ 8x + 15 = 0 giving factor (x+3)(x+5) rational
m = 2 gives (3,-16, -35) we get 3x^2 -16x -35= 0 giving (3x+5)(x-7)
m =3 gives (15,-24,-63) divide by 3 to get 5x^2-8x-21 = (5x+7)x-3)

so on
 
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Back
Top