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
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Is it possible to arrange six pencils such that each one touches the other five? If so, how? This is an adaption of a Martin Gardner puzzle only I changed it from cigarettes to pencils and left out the clues because PF folks don’t need clues. From the book “My Best Mathematical and Logic Puzzles”. Dover, 1994.
Thread 'Imaginary Pythagoras'
I posted this in the Lame Math thread, but it's got me thinking. Is there any validity to this? Or is it really just a mathematical trick? Naively, I see that i2 + plus 12 does equal zero2. But does this have a meaning? I know one can treat the imaginary number line as just another axis like the reals, but does that mean this does represent a triangle in the complex plane with a hypotenuse of length zero? Ibix offered a rendering of the diagram using what I assume is matrix* notation...
Back
Top