# Find $K$ in the equation $24x^3-14x^2-63x+K=0$ if one root is twice more the another.

I am given the equation $24x^3-14x^2-63x+K=0$ and I am asked to find the values for $K$ If one root is double the other root.

How would I solve this? I am trying to solve this by taking two roots as A,2A and the third B then by using sum and products of roots I wrote three equations.

• did you try dividing by $24$ and using vietta? – Jorge Fernández Hidalgo Aug 1 '16 at 18:25
• @CarryonSmiling what is vietta? – danny Aug 1 '16 at 18:28
• wikiwand.com/en/Vieta's_formulas – alans Aug 1 '16 at 18:28
• I am thinking of solving this by using sum and product of roots – danny Aug 1 '16 at 18:29
• @CarryonSmiling, Perhaps Danny is an algebra student from secondary school. I highly doubt they would be required to know Vietta's formulas. – KingDuken Aug 1 '16 at 18:38

By using Viete's formulae for roots $x$, $2x$, $y$ of equation $24x^3-14x^2-63x+K=0$, we get system of equations $$3x+y=\frac{7}{12},$$ $$3xy+2x^2=-\frac{21}{8},$$ $$K=-48x^2y.$$ From the first equation $y=\frac{7}{12}-3x$, substitute this into second equation in order to get $8x^2-2x-3=0$. Now it is easy to conclude that solutions are $$(x,y)\in\{(\frac{3}{4},-\frac{5}{3}),(-\frac{1}{2},\frac{25}{12})\}.$$ Finally, calculate $K$ from the third equation, for those values $x$ and $y$: $$K\in\{45,-25\}.$$

• The quadratic equation obtained by substituting the first equation into the second is actually $7x^2 - \frac{7x}{4}-\frac{21}{8} = 0$. The roots you obtained are correct however. – Joey Zou Aug 1 '16 at 21:09
• Yes, I multiplied that equation with $\frac{8}{7}$ in order to get 'more beautiful' equivalent equation, roots didn't change by doing this. – alans Aug 1 '16 at 21:13
• Ah, of course, silly me. – Joey Zou Aug 1 '16 at 21:14
• @Joey We can get even "more beautiful" equations be eliminating fractions as much as possible, right from the start - see my answer. – Bill Dubuque Aug 2 '16 at 0:12

The solution is computable with simple mental arithmetic if we eliminate fractions by scaling the polynomial by $72$ to transform it to have leading coeff $=1\$ (AC method).  This yields

$$\quad\ X^3 - 7 X^2 -63\cdot 6 X + 72K = 0,\ \ X = 12x$$

Consider more generally $\ X^3 - a\, X^2 + b\, X + c = 0\$ with roots $\,X,\,2X,\, Y.\$ By Vieta

$$3X+Y = a,\ \ 3XY+2X^2 = b$$

Eliminate $Y$ and scale by $\,7\,$ to get $\ (7X)^2 - 3a (7X) + 7b = 0\$ with solution

$$7X = (3a\pm d)/2,\ \ d = \sqrt{9a^2-28b}\qquad$$

In OP $\,\ 9a^2\!-28b = 9(7)^2\!+4\ 7\ (3^2\ 7\ 6) = 3^2 7^2 (1\!+\!24) =(3\ 7 \ 5)^2 = 105^2$

Therefore $\ X = (21\pm 105)/14 = 9,\, -6,\ \$ so $\ \ Y = 7-3X = -20,\, 25$

which  implies $\ \ x = X/12 = 3/4,\, -1/2\ \$ and $\ \ y = Y/12 = -5/3,\, 25/12$

Suppose $x^3-\frac{7}{12}x^2-\frac{63}{24}x+\frac{K}{24}=(x-a)(x-2a)(x-b)$.

So $\frac{63}{24}=2a^2+3ab$ and $\frac{7}{12}=-3a-b$.

We conclude $b=-3a-\frac{7}{12}$.

So we must solve $\frac{63}{24}=2a^2-3a(3a+\frac{7}{12})$.

From here you can find $a$, and once you have $a$ you get $b$.

Finally $K=24(-2a^2b)$

• I still think they might be a form to deduce $K$ without getting determining $a$ and $b$, because they turn out to be complex. – Jorge Fernández Hidalgo Aug 1 '16 at 18:41
• a and b are not complex as you can see from my answer, you made a mistake-factor by x is $\frac{21}{8}$ and I think you didn't apply Viete's formulae regularly. – alans Aug 1 '16 at 19:11
• @CarryonSmiling To avoid mistakes with fractions we can eliminate them! See my answer for one way to do that. – Bill Dubuque Aug 2 '16 at 0:14