Factorize $abx^2-(a^2+b^2)x+ab$

Question

Factorize the quadratic trinomial $$abx^2-(a^2+b^2)x+ab.$$ The discriminant of the trinomial is $$D=(a^2+b^2)^2-(2ab)^2=\\=(a^2+b^2-2ab)(a^2+b^2+2ab)=(a-b)^2(a+b)^2=\\=\left[(a-b)(a+b)\right]^2=(a^2-b^2)^2\ge0 \text{ } \forall a,b.$$ So the roots are $$x_{1,2}=\dfrac{a^2+b^2\pm\sqrt{(a^2-b^2)^2}}{2ab}=\dfrac{a^2+b^2\pm\left|a^2-b^2\right|}{2ab}.$$ How can I expand the modulus here? Have I calculated the discriminant in a reasonable way? Can we talk about "the discriminant of a quadratic trinomial" or only the corresponding quadratic equation (the trinomial=0) has a discriminant? What about "the roots of a trinomial" (or of the corresponding quadratic equation)?

The quadratic is symmetric in $a, b$ i. e replacing $a$ with $b$ and $b$ with $a$ gives the same quadratic so without losing generality you can assume $|a|>|b|$ which implies $a^2>b^2$, so you can safely assume $a^2>b^2$. — kingW3, Apr 07 '21 at 11:51

score 6 · Accepted Answer · answered Apr 07 '21 at 11:05

6

Using the quadratic formula is a good approach, and everything you have done so far is correct. The modulus/absolute value can be replaced by normal brackets as there is a $\pm$ preceding it, and the numbers involved here are real. Notice that:

$$\frac {a^2 +b^2 + (a^2 - b^2)}{2ab} = \frac ab, \quad \frac {a^2 +b^2 - (a^2 - b^2)}{2ab} = \frac ba$$

hence the roots are $\dfrac ab$ and $\dfrac ba$, and we are looking at the factors $(bx- a)$ and $(ax - b)$. Now:

$$(bx-a)(ax- b) = abx^2 - a^2 x - b^2 x + ab$$

which is exactly what we are trying to factorize.

answered Apr 07 '21 at 11:05

player3236

16,600

Thank you for the response! May I ask you why when we have $\pm$ sign before an absolute value, we can replace it by normal brackets? I don't see it straight-forward. – Math Student Apr 07 '21 at 11:15
If $x \ge 0$, $\pm |x| = \pm x$. If $x \le 0$, $\pm |x| = \pm (-x) = \mp x$, which is equal to $\pm x$ in this context (there is only one appearance of $\pm/\mp$.) So $\pm |x| = \pm x$. – player3236 Apr 07 '21 at 11:20
Thank you for the explanation. Is the reason for $\mp$ to be equal to $\pm$ in our context the fact that there is actually only one appearance of $\pm$ or $\mp$? What will be the problem if there were two? – Math Student Apr 07 '21 at 11:25
2

Usually if there are two or more $\pm$ or $\mp$s in a formula, we read it as two formulas merged into one. For example, we have $$\tan (x\pm y) = \frac {\tan x \pm \tan y}{1 \mp \tan x \tan y}$$ In these cases if there is an absolute value sign, it is safer to consider the cases where they are positive or negative seperately. – player3236 Apr 07 '21 at 11:28

Dietrich Burde · Answer 2 · 2021-04-07T11:12:48.193

4

For the title question. The factorization is given by $$ (ax-b)(bx-a)=abx^2-(a^2+b^2)x+ab. $$ We can find it by multiplying out $(\rho_1x+\rho_2)(\rho_3x+\rho_4)$ and comparing coefficients. This is easier than computing a discriminant.

edited Apr 07 '21 at 11:12

answered Apr 07 '21 at 11:08

Dietrich Burde

140,055

1

You have the wrong sign on the x coefficient – Ben Apr 07 '21 at 11:11
@Ben Ooops, thank you - corrected. – Dietrich Burde Apr 07 '21 at 11:13

score 3 · Answer 3 · answered Apr 07 '21 at 11:28

Using the Quadratic Formula is more general but here is an alternative method:

We have $abx^2-(a^2+b^2)x+ab=0 \iff x^2-\frac{a^2+b^2}{ab}x+1=0 \iff x^2-(\frac{a}{b}+\frac{b}{a})x+1=0$.

Let $r_1$ and $r_2$ be the (possibly equal, possibly complex) roots of the quadratic function $x^2-(\frac{a}{b}+\frac{b}{a})x+1$. Then their product is $1$ and their sum is $\frac{a}{b}+\frac{b}{a}$. This shows that the roots are $\frac{b}{a}$ and $\frac{a}{b}$.

Therefore, $x^2-(\frac{a}{b}+\frac{b}{a})x+1=(x-\frac{a}{b})(x-\frac{b}{a})$ so

$$abx^2-(a^2+b^2)x+ab = ab(x-\frac{a}{b})(x-\frac{b}{a}) = (ax-b)(bx-a)$$

lone student · Answer 4 · 2021-04-09T22:56:22.487

2

Everything is OK.

First: Accept

$$a^2≥b^2$$

Then accept

$$a^2≤b^2$$

You will see that the roots have not changed.

We have

$$abx^2-(a^2+b^2)x+ab=(a x - b) (b x - a)$$

Small Supplement:

Note that,

$$\begin{cases} ±|a^2-b^2|=±(a^2-b^2), a^2≥b^2 \\ ±|a^2-b^2|=±(b^2-a^2), a^2≤b^2 \end{cases}$$

$$\implies \pm(a^2-b^2)=\mp(b^2-a^2)$$

Thus,

$$±|a^2-b^2|=±(a^2-b^2).$$

edited Apr 09 '21 at 22:56

answered Apr 07 '21 at 11:06

lone student

1

Thank you for the response! You are trying to show me that $\pm|A|=\pm(A)$, right? – Math Student Apr 07 '21 at 11:18
2

@Medi Sure. I tried to answer your comment. If you still have questions, feel free to ask. – lone student Apr 07 '21 at 11:25

score 0 · Answer 5 · answered Jan 05 '22 at 07:16

The polynomial $ \ Cx^2 + Bx + A \ $ is referred to as the reciprocal polynomial to $ \ Ax^2 + Bx + C \ \ , $ as its zeroes are the reciprocals of the latter polynomial. In the same vein, a polynomial $ \ Ax^2 + Bx + A \ $ is called palindromic and has the property that if $ \ r \ $ is a zero, then $ \ \frac{1}{r} \ $ is as well. (This applies to polynomials of higher degree as well as quadratics.)

This permits us to write $$ ab·x^2 \ - \ (a^2+b^2)·x \ + \ ab \ \ = \ \ ab · \left[ \ x^2 \ - \ \frac{a^2+b^2}{ab}·x \ + \ 1 \ \right] $$ $$ = \ \ ab · \left[ \ x^2 \ - \ \left(\frac{b}{a } + \frac{a}{b}\right)·x \ + \ 1 \ \right] $$ $$ = \ \ ab·(x - r)· \left(x - \frac{1}{r} \right) \ \ = \ \ ab·[x^2 - \left(r + \frac{1}{r} \right)·x \ + 1] \ \ . $$

The reciprocal nature of the two zeroes is thus confirmed and the factorization proceeds as in Taladris's answer.

Factorize $abx^2-(a^2+b^2)x+ab$

5 Answers5