If the given quadratic equation has two integral roots $\alpha$ and $\beta$ then find $\alpha + \beta$ and $|\alpha-\beta|$

Question

The following is the given quadratic equation:$$x^2-\frac{(\alpha ^2+11)x}{9}+\frac{15}{4}(\alpha +\beta ) +16=0.$$

I used Vieta’s formula $$\frac{(\alpha)^2+11}{9}=\alpha+\beta$$ $$\alpha\times\beta=\frac{15}{4}(\alpha+\beta)+16$$

From the above equation I get a cubic $$4\alpha^3-204\alpha^2+44\alpha-741=0$$

And I got stuck.

Is there any way to find directly the value of $\alpha + \beta$?

Answer 1

Your cubic formulation is incorrect. The correct cubic equation is $$4x^3 - 51x^2 + 44x - 741 = 0$$

Unfortunately, I could not find a better way of solving this problem other than bruteforcing this cubic using the rational root theorem. 13 is the only real root of this cubic, which implies $\alpha = 13$. Substituting this, you can easily find that $\beta = 7$, which means $\alpha + \beta = |\alpha + \beta| = 20$

For a quick check, substitute $\alpha=13$ and $\beta=7$ in the quadratic to get $$x^2 - 20x + 91 = 0$$ which has roots $7$ and $13$, proving that our solution is correct.

Answer 2

I'm just gonna replace $\alpha$ with $x$ and $\beta$ with $y$ to make it easier to type: $$\frac{x^2+11}{9}=x+y$$ $$xy=\frac{15}{4}(x+y)+16$$ We can assume WLOG that $x>y$ let $a=x+y$ and $b=x-y$ so we get $$\frac{\left(\frac{a+b}{2}\right)^2+11}{9}=a \tag{1}$$ $$\frac{a+b}{2}\cdot\frac{a-b}{2}=\frac{15}{4}a+16 \tag{2}$$ $$(2) \implies a^2-b^2=15a+64$$ $$(1) \implies a^2+2ab+b^2+44=36a$$ Subtracting the second equation from the first we get $$2b^2+2ab+44=21a-64 \iff 2b^2+108=a(21-2b)$$ and adding them we get $$2a^2+2ab+44=51a+64 \iff 2a^2+(2b-51)a-20=0$$ I think it shouldn't be too hard to continue from here. (try letting $z=2b-21$ and getting rid of all $b$'s and see what happens)

Answer 3

In our answer given below, we address the point at issue, i.e. solving a cubic equation, which has stonewalled your attempt to determine the value of $\alpha$. Our aim is to make you aware of a method appropriate for all seasons. Let us start by stating the general cubic equation in its $\mathrm{normal\space form}$ as $$ax^3+bx^2+cx+d=0 \tag{1}.$$ By substituting $$x=y-\frac{b}{3a} \tag{2}$$ in equation (1), we can express it in its $\mathrm{reduced\space form}$ as $$y^3+3py+2q=0, \quad\mathrm{where} \tag{3}$$ $$p=\frac{c}{3a}-\frac{b^2}{9a^2} \quad\mathrm{and} \tag{4}$$ $$q=\frac{b^3}{27a^3}-\frac{bc}{6a^2}+\frac{d}{2a}. \tag{5}$$

We need another auxiliary value $r$, which is given by the equation below. $$r=\pm\sqrt{\vert p\vert} \tag{6}$$

As you can see from equation (6), the value of $r$ depends solely on the value of $p$. However, when it comes to deciding the sign of $r$, $q$ calls the shots, i.e. the sign of $r$ is the same as that of $q$.

If we want to know the type of roots of equation (3), we need to calculate the discriminator $D$ using the following equation. $$D=q^2+p^3 \tag{7}$$

Using $D$, we can state :

$\bullet\space$ for $D \gt 0$, 1 real root and 2 complex roots,

$\bullet\space$ for $D \lt 0$, 3 different real roots,

$\bullet\space$ for $D=0$, 3 equal real roots (if $p=q=0$) or 3 real roots, where 2 are equal (if $p^3=-q^2\neq 0$).

Now we have to consult the $\mathrm{Table 1}$ in order to determine the three roots $y_1$, $y_2$, and $y_3$ of the cubic equation given in equation (3). When we have them, we can substitute them in equation (2) to determine the three roots of equation (1).

In your case, $a=4$, $b=-51$, $c=44$, and $d=-741$. Please note that the equation you have posted is wrong, because you have forgotten to divide the coefficient of $\alpha^2$ by 4. We obtain the following values for $p$, $q$, and $r$ using equations (4), (5), and (6) respectively.. $$p=-\frac{691}{48}\qquad q=-\frac{9345}{64}\qquad r=-3.7941841459440701918596606235852$$ Therefore, according to equation (7), $D= 18337.169994212962962962962962963\gt 0$ - meaning that equations (1) and (3) has only one real roots. We also noticed that $p\lt 0$. Therefore, the sought roots are given in the $2^{nd}$ column of the $\mathrm{Table 1}$. Continuing our calculation, we obtain, $$\cosh\left(\phi\right)=\frac{q}{r^3}=2.6732779532138537322855297425166,$$ $$\therefore \quad\phi=1.6394770650491883793707521746177,$$ $$\space\therefore \quad\frac{\phi}{3}=0.54649235501639612645691739153923,$$ $$\qquad\quad\space\therefore \quad\cosh\left(\frac{\phi}{3}\right)= 1.1530805653375611292192890462415.$$ The only real root $y_1$ is determined as $$y_1=-2\times 3.7941841459440701918596606236\times 1.153080565337561129219289046 =8.75.$$ We use equation (2) as shown below to find the value of $\alpha$. $$\alpha=y_1-\frac{b}{3a}=8.75-\left(\frac{-51}{12}\right)=13$$

Answer 4

As we are told that the roots $ \ \alpha \ $ and $ \ \beta \ $ are integers, your Viete relations $$ \frac{\alpha^2 \ + \ 11}{9} \ = \ \alpha \ + \ \beta \ \ \ , \ \ \ \alpha·\beta \ = \ \frac{15}{4}·(\alpha \ + \ \beta) \ + \ 16 $$ indicate that there are important constraints on the value of these roots.

Since the roots are integers, their sum and product must be integers as well. The first equation then tells us that $ \ a^2 + 11 \ $ is an integer multiple of $ \ 9 \ \ . $ The sum of the roots appears in the second equation, where it must be evenly divisible by $ \ 4 \ $ in order that the product of the roots be an integer. Thus, $ \ a^2 + 11 \ $ must be a positive integer multiple of $ \ 36 \ \ . $ This in turn means that the product of the roots is positive.

The discriminant of the quadratic polynomial we seek is $$ \Delta \ \ = \ \ (\alpha \ + \ \beta)^2 \ - \ 4·1·[\frac{15}{4}·(\alpha \ + \ \beta) \ + \ 16] \ \ = \ \ (\alpha \ + \ \beta)^2 \ - \ 15·(\alpha \ + \ \beta) \ - \ 64 \ \ . $$ The sum of the roots is real then for $ \ \Delta \ \ge \ 0 \ \Rightarrow \ (\alpha \ + \ \beta) \ \le \ \approx -3.47 \ \ $ or $ \ \ (\alpha \ + \ \beta) \ \ge \ \approx 18.47 \ \ . $ The sum of the roots must therefore be an integer greater than or equal to $ \ 19 \ \ . $

To resolve the value(s) of the roots, it will be useful to look at the squares of the integers modulo $ \ 9 \ \ , $ which are $ \ 0 \ , \ 1 \ , \ 4 \ , \ 0 \ , \ 7 \ , \ 7 \ , \ 0 \ , \ 4 \ , \ 1 \ \ . $ With $ \ 11 \ \equiv \ 2 \ [\mod 9 \ ] \ \ , $ we find $ \ \alpha^2 + 11 \ \equiv \ 0 \ [\mod 9 \ ] \ \ $ for $ \ \alpha \ \equiv \ 4 \ , \ 5 \ [\mod 9 \ ] \ \ . $

Since $ \ \alpha^2 + 11 \ $ must also be a positive multiple of $ \ 4 \ \ , $ the first candidate is $ \ \alpha \ = \ 5 \ \Rightarrow \ 5^2 + 11 \ = \ 36 \ \ . $ But this produces $ \ \alpha + \beta \ = \ 4 \ \ , $ which is too small. The next largest candidate is $ \ \alpha \ = \ 13 \ \Rightarrow \ 13^2 + 11 \ = \ 180 \ \ . $ From this, we obtain $ \ \alpha + \beta \ = \ 20 \ \Rightarrow \ \beta \ = \ 7 \ \ , $ as found in the other posted answers. [We also note that $ \ |\alpha \ - \ \beta| \ = \ 6 \ \ . ] $

The candidates beyond this are $ \ 23^2 + 11 \ = \ 540 \ \ $ and $ \ 31^2 + 11 \ = \ 972 \ \ , $ which do not produce consistent results for $ \ \alpha·\beta \ \ . $ It is at this point that we can make use of the cubic polynomial you obtained by combining the Viete relations, $ \ 4·\alpha^3 \ - \ 51·\alpha^2 \ + \ 44·\alpha \ - \ 741 \ \ . $ Having established that $ \ \alpha = 13 \ $ is a zero, we find the factorization $ \ (\alpha - 13) \ · \ (4·\alpha^2 \ + \ \alpha \ + \ 57) \ \ , $ so there are plainly no other real zeroes. Hence, the solution roots $ \ \mathbf{\alpha \ = \ 13 \ , \ \beta \ = \ 7} \ $ are unique.