Show that ##(b-c)x^2+(c-a)x+a-b=0## has rational roots

[chwala]

Homework Statement

Show that $(b-c)x^2+(c-a)x+a-b=0$ has rational roots

Relevant Equations

discriminant

If we have a quadratic equation, $px^2+qx+d$ ,then the condition that the roots are rational is satisfied if our discriminant has the form $q^2-4pd≥0$ (also being a perfect square). Therefore we shall have,
$(c-a)^2-4(b-c)((a-b)≥0$
$(c-a)^2-4(ab-b^2-ac+bc)≥0$
$(c-a)^2-4[b(a-b)-c(b-a)]≥0$
$(c-a)^2+4[b(b-a)-c(b-a)]≥0$also...
$(c^2+a^2+4b^2+4ac)-4bc-2ac≥0$... i may need to analyse this later...i hope i am on the right track...i was also thinking of using the complete the square approach to prove this...

 

[anuttarasammyak]

The equation is written as
$$(x-1)[(c-b)x+(a-b)]=0$$
I assume the problem states something about a,b,c.

 

[PeroK]

If $b^2-4ac≥0$ is positive and a perfect square then the roots are rational.

This is wrong, as was pointed out in your previous thread. You can't use the same letters for different things in the same question.

 

[chwala]

The equation is written as
$$(x-1)[(c-b)x+(a-b)]=0$$
I assume the problem states something about a,b,c.

That is not stated...i have typed exactly as it appears on my textbook. see attached;

old textbook ...first published in 1973...

 

[chwala]

This is wrong, as was pointed out in your previous thread. You can't use the same letters for different things in the same question.

I will amend and use different variables...talk later...

 

[anuttarasammyak]

Now we get two solutions. The solution other than 1 is rational or not depends on what are a,b and c.

 

[anuttarasammyak]

Homework Statement:: Show that $(b-c)x^2+(c-a)x+a-b=0$ has rational roots
Relevant Equations:: discriminant

If we have a quadratic equation, $px^2+qx+d$ ,then the condition that the roots are rational is satisfied if our discriminant has the form $q^2-4pd≥0$ (also being a perfect square). Therefore we shall have,

The roots are
$$x=\frac{-q \pm \sqrt{q^2-4pd}}{2p}$$
where
$$p=b-c, \ q=c-a, \ d=a-b$$
so
$$x=\frac{-(c-a) \pm \sqrt{(a-c)^2-4(b-c)(a-b)}}{2(b-c)}=\frac{-(c-a) \pm (2b-a-c)}{2(b-c)}$$
Thus we confirm #2.

 

[chwala]

The roots are
$$x=\frac{-q \pm \sqrt{q^2-4pd}}{2p}$$
where
$$p=b-c, \ q=c-a, \ d=a-b$$
so
$$x=\frac{-(c-a) \pm \sqrt{(a-c)^2-4(b-c)(a-b)}}{2(b-c)}=\frac{-(c-a) \pm (2b-a-c)}{2(b-c)}$$
Thus we confirm #2.

I guess i may be tired...how did you factorise $a^2+2ac+c^2+4(b^2-ab-bc)?$ into a perfect square? Indeed its true that if you expand lhs we get the rhs... $(2b-a-c)^2=a^2+2ac+c^2+4(b^2-ab-bc)$ ... wah

 

[chwala]

The roots are
$$x=\frac{-q \pm \sqrt{q^2-4pd}}{2p}$$
where
$$p=b-c, \ q=c-a, \ d=a-b$$
so
$$x=\frac{-(c-a) \pm \sqrt{(a-c)^2-4(b-c)(a-b)}}{2(b-c)}=\frac{-(c-a) \pm (2b-a-c)}{2(b-c)}$$
Thus we confirm #2.

Thanks...you made it look easy anutta...The key in this question ( been trying to figure that out)... was to try and express the discriminant as a perfect square. Cheers man!

 

[PeroK]

It's a lot easier if you notice that $1$ is always a root of the quadratic:
$$(b-c)1^2+(c-a)1+a-b = b - c + c - a + a - b = 0$$That allows you to factorise as in post #2:

$$(x-1)[(b-c)x-(a-b)]=0$$

And the other root is $\frac{a -b}{b-c}$.

 

[chwala]

It's a lot easier if you notice that $1$ is always a root of the quadratic:
$$(b-c)1^2+(c-a)1+a-b = b - c + c - a + a - b = 0$$That allows you to factrorise as in post #2:

And the other root is $\frac{a -b}{b-c}$.

wawawawawa you guys are smart...thanks Perok i learned something new today (i.e 1 always being a factor of a quadratic. bingo!)...therefore,
if $f(x)= bx^2-cx^2+cx-ax+a-b$ and we have the factor $x-1$, then i have just checked using long division that,
$f(x)= bx^2-cx^2+cx-ax+a-b≡(x-1)(bx-cx+b-a)$ wah!

 

[PeroK]

wawawawawa you guys are smart...thanks Perok i learned something new today (i.e 1 always being a factor of a quadratic. bingo!).

Not any quadratic, only this one.