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I am stuck with this equation, I need to prove the roots are real when $a, b, c \in R$

The equation is $$(a+b-c)x^2+ 2(a+b)x + (a+b+c) = 0$$

If someone could tell me the right way to go about this, so I can attempt it.

Thank you

EDIT: I have made an error in the question. I have now corrected it.

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    $\begingroup$ Do you know the relationship between the discriminant of a polynomial and it's roots? $\endgroup$ – Tom Oldfield Sep 3 '13 at 19:19
  • $\begingroup$ You must have misread the question. The roots are only real if $|a+b+c| \le |a+b|$. $\endgroup$ – TonyK Sep 3 '13 at 19:29
  • $\begingroup$ I had an error in the question, it is now corrected $\endgroup$ – user Sep 3 '13 at 19:33
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We look at the discriminant of the the polynomial, which for a quadratic $ax^2 +bx +c$ is $b^2 -4ac$, plugging the values in for our polynomial gives $$\Delta = 4(a+b)^2-4(a+b-c)(a+b+c)\\ = 4[(a+b)^2 - (a+b)^2+c^2]\\ = 4c^2$$

Since the square of a real number is positive, we know that the roots must be real, by looking at the quadratic forumla and seeing that the solutions are $$\frac{-b\pm\sqrt\Delta}{2a}$$

And the square root of a positive real number is real. We used the discriminant because it makes computation so much easier, than if we were doing everything that we did in the first step underneath the radical, and it would be rather ugly. Inspection shows that if $\Delta > 0$, there are two distinct real roots, if $\Delta < 0$, there are two complex roots, which are conjugate, and if $\Delta = 0$ then you have a real double root.

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  • $\begingroup$ Sorry, typo in question. $\endgroup$ – user Sep 3 '13 at 19:35
  • $\begingroup$ All fixed with the new problem! $\endgroup$ – guest196883 Sep 3 '13 at 19:41
  • $\begingroup$ we know that the roots must be positive - probably a typo, they must be real. $\endgroup$ – Džuris Sep 3 '13 at 19:45
  • $\begingroup$ Yes yes, thanks. It was a typo. $\endgroup$ – guest196883 Sep 3 '13 at 19:46
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    $\begingroup$ All answers were really good but this one just that bit clearer. Thanks $\endgroup$ – user Sep 3 '13 at 21:01
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Set $a+b-c=A$ and $a+b+c=B$; then $2(a+b)=A+B$ and your equation is $$ Ax^2+(A+B)x+B=0 $$ that can be written $$ Ax^2+Ax+Bx+B=0 $$ or $$ Ax(x+1)+B(x+1)=0 $$ and finally $$ (Ax+B)(x+1)=0. $$ If $A=0$, then the equation is $B(x+1)=0$. If also $B=0$, the equation is an identity; otherwise it has the only solution $-1$. If $A\ne0$, the solutions are $-1$ and $-B/A$.

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Note that $a$ and $b$ only appear in the expression $a+b$. So let us simplify our lives and denote $a+b =:d $, so we need to show that the following has real roots:

$$ (d-c)x^2 + 2dx + (c+d) = 0.$$

There is a useful object called the discriminant. For an equation $Ax^2 + Bx + C$ it is given as: $\Delta = B^2 - 4 AC$. The roots of a polynomial are real iff $\Delta \geq 0$. For our polynomial, we have:

$$\Delta = 4d^2 - 4(d+c)(d-c) = 4 c^2 \geq 0. $$

Because $\Delta \geq 0$, the roots are real, and you are done.


By the way, the roots of $Ax^2 + Bx + C$ are $\frac{-B \pm \Delta}{2A}$, which is very often very useful.

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  • $\begingroup$ Typo in the question, Sorry $\endgroup$ – user Sep 3 '13 at 19:35
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$$ax^2 +bx^2 + cx^2 + 2ax+2bx + a+b+c = 0$$

or

$$x^2(a+b-c) + 2x(a+b) +a+b+c = 0$$

Let us now find the expression for the discriminant,

$$\Delta = 4(a+b)^2-4(a+b-c)(a+b+c)$$

If we prove that, $\Delta\geq 0$, we done.

$$4(a+b)^2-4(a+b+c)(a+b-c)=$$

$$=4a^2+8ab+4b^2-4(a^2+2ab+b^2-c^2)=$$

$$=4a^2+8ab+4b^2-4a^2-8ab-4b^2+4c^2=$$

$$4c^2\geq 0$$

Hence, indeed

$$\Delta\geq 0$$

and therefore the quadratic equation has real roots.

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You could note that $x=-1$ is a root.

If this is not obvious at first, write (motivated by the fact that $a,b$ always appear as $a+b$)

$$0=(a+b-c)x^2+2(a+b)x+(a+b+c)=(a+b)(x+1)^2-c(x^2-1)$$$$=(x+1)\left((a+b)(x+1)-c(x-1)\right)=(x+1)\left((a+b-c)x+(a+b+c)\right)$$

Note that if $a+b=c$ the equation is linear (just one root), and that if $c=0, a+b\neq 0$ there is a double root at $x=-1$

Obviously going via the discriminant is a reliably effective way. With coefficients like these it is worth testing $x=\pm1$ just in case it saves a lot of work.

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