Characterizing Cubics!

Algebra Level 5

f ( x ) = x 3 + p x + q \large{f(x) = x^3 + px + q}

Let f ( x ) = 0 f(x) = 0 be a cubic equation such that;

  • it has one repeated root of multiplicity 2, and

  • p p and q q are integers such that q = n p q=np for the least permissible positive integer n n .

Find the value of p + q p+q .


The answer is -81.

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Satyajit Mohanty by Satyajit Mohanty , 24, India

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3 solutions

Let the repeated root be a a and the third root b . b. Then

f ( x ) = ( x − a ) ( x − a ) ( x − b ) = x 3 − ( 2 a + b ) x 2 + ( a 2 + 2 a b ) x − a 2 b . f(x) = (x - a)(x - a)(x - b) = x^{3} - (2a + b)x^{2} + (a^{2} + 2ab)x - a^{2}b.

Comparing like coefficients to f ( x ) = x 3 + p x + q f(x) = x^{3} + px + q informs us that

  • 2 a + b = 0 ⟹ b = − 2 a , 2a + b = 0 \Longrightarrow b = -2a,

  • a 2 + 2 a b = p ⟹ a 2 + 2 a ( − 2 a ) = − 3 a 2 = p a^{2} + 2ab = p \Longrightarrow a^{2} + 2a(-2a) = -3a^{2} = p and

  • − a 2 b = q ⟹ − a 2 ( − 2 a ) = 2 a 3 = q . -a^{2}b = q \Longrightarrow -a^{2}(-2a) = 2a^{3} = q.

Now if a = 0 a = 0 then we would have b = 0 , b = 0, so given that the repeated root has a multiplicity of precisely 2 2 we must have a ≠ 0. a \ne 0.

Now with a 2 = − p 3 a^{2} = -\dfrac{p}{3} we have that q = 2 a ∗ a 2 = − 2 3 a p . q = 2a*a^{2} = -\dfrac{2}{3}ap.

Thus to make q q the least possible positive integral multiple of p p we choose a = − 3 , a = -3, giving us q = 2 p . q = 2p.

Then with a = − 3 a = -3 we find that p = − 3 a 2 = − 27 p = -3a^{2} = -27 and q = 2 p = − 54 , q = 2p = -54, for which p + q = − 81 . p + q = \boxed{-81}.

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Satyajit Mohanty Nice question. I think that for sake of clarity you might want to rephrase the first condition as "it has one repeated root of multiplicity 2 2 " to rule out the possibility of having 0 0 as a root of multiplicity 3. 3.

Brian Charlesworth - 5 years, 9 months ago

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Thanks for pointing that out. I've edited the problem for clarity. And, your solution is Brilliant.

Satyajit Mohanty - 5 years, 9 months ago

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Thank you! :)

Brian Charlesworth - 5 years, 9 months ago

@Satyajit Mohanty @Brian Charlesworth I think the statement "q is the smallest permissible positive integral multiple of p" should be improved for clarity. I thought that q is the smallest positive integral value which is a multiple of p. And consequently I got p=-27 and q=54, giving p+q=27.

BTW Nice Problem with a nice solution!

Sandeep Bhardwaj - 5 years, 9 months ago

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Yes, I can see how it could be interpreted that way. Would something like " p p and q q are integers such that q = n p q = np for the least permissible positive integer n n " provide clarity?

Brian Charlesworth - 5 years, 9 months ago

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Yeah it will!

Sandeep Bhardwaj - 5 years, 9 months ago

Thanks. I've amended the problem :)

Satyajit Mohanty - 5 years, 8 months ago

the second statement didnt seem to say n to be an integer, but a positive

Aareyan Manzoor - 5 years, 8 months ago

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You're right, it really should read as "positive integer n n ". A previous phrasing of the question made it clear that n n must be integral, while now it is implicit, (since the letter n n is usually reserved for integers). Perhaps Satyajit may choose to add the word "integer" to make that condition more explicit.

Brian Charlesworth - 5 years, 8 months ago

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i assumed n not to be integral and gave answer 0,sir.should i be marked?

Aareyan Manzoor - 5 years, 8 months ago

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@Aareyan Manzoor – Referring to my solution, in order to have p + q = 0 p + q = 0 we would require that

− 3 a 2 + 2 a 3 = 0 ⟹ a 2 ( 2 a − 3 ) = 0 , -3a^{2} + 2a^{3} = 0 \Longrightarrow a^{2}(2a - 3) = 0, in which case either a = 0 a = 0 or a = 3 2 . a = \dfrac{3}{2}.

If a = 0 a = 0 then since 2 a + b = 0 2a + b = 0 we would require that b = 0. b = 0. But since by the first condition not all three roots can be the same, we cannot have a = 0. a = 0.

If a = 3 2 a = \dfrac{3}{2} then p = − 3 a 2 = − 27 4 p = -3a^{2} = -\dfrac{27}{4} and q = 2 a 3 = 27 4 , q = 2a^{3} = \dfrac{27}{4}, which would give us q = − p q = -p and thus n = − 1 , n = -1, which is not positive.

So an answer of 0 0 would not be consistent with the condition that n n be positive, regardless of whether or not it is an integer. So unless I'm mistaken, I don't think an answer of 0 0 can be considered as correct. Sorry. :( Did you try any values other than 0 0 on your three attempts?

Brian Charlesworth - 5 years, 8 months ago

Discriminat must be zero because of a root has multiplicity of 2. ∴ − 4 ∗ p 3 − 27 q 2 = 0. B u t q = n ∗ p , ⟹ − 4 ∗ p 3 = 27 ∗ n 2 ∗ p 2 , b u t p ≠ 0. ∴ p < 0 , a n d n 2 = − 4 ∗ p 27 . B u t p a n d n a r e i n t e g e r s , n > 0. ∴ m i n i m u m v a l u e o f n w o u l d b e w h e n p = − 27 , n = 2. ⟹ q = 2 ∗ ( − 27 ) = − 54. p + q = − 81 \therefore~ - 4*p^3 - 27q^2=0.~~But~q=n*p,~~\implies~- 4*p^3=27*n^2*p^2,~ but~p\neq0.\\ \therefore~p<0,~and~n^2= - \dfrac{4*p}{27}.~~~But~p~and~n~are~integers, ~n>0.\\ \therefore~minimum~value~ of~ n~ would~ be~ when~ p= - 27, n=2.~~ \implies q= 2*( - 27)= - 54.\\ p+q= \Large~~~\color{#D61F06}{ - 81}\\~~\\

Just an addition. Since cof. of x square is zero, and a root with multiplicity 2, we can say that the roots are r, r, -2r, and − ( − 54 ) = 2 ∗ r 3 . ∴ -( - 54)=2*r^3.~~\therefore the roots are -3, -3, 6. Note that q is twice the cube of a number.
I use TI-83 calculator. For graphing window[ -4,7] in X, with [-1,1] I got graph touching x-axis. But Y-window [ - 0.1, 0.1], it showed the graph is not touching. If it is an intersection, 1 E -10 gives very good result, but be careful when it is touching graph. In fact this problem also was first solved in TI-83 by graphing:- X 3 + p ∗ ( X + n ) , X^3+p*( X+n), and setting p+1-->p. With - 30<X<30 . We can see with increase in n, the curve moving down. Only 1<= n<=3 was needed.

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Ruslan Abdulgani
Sep 18, 2015

f(x) = (x – a)^2(x+(np)/a2) = x^3 + [(np/a^2) – 2a]x^2 + (a^2 – 2np/a) x +np. From (np/a^2) – 2a = 0, we get p=2a^3/n, and from a^2 – 2np/a = p, - 3a^2 = 2a^3/n, n=2a/-3, smallest n=2, for a=-3.So p=-27. f(x)=(x+3)2(x – 6) = x3 – 27x – 54 =0, p+q= -27 – 54 = -81

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f ( x ) = ( x − a ) 2 ( x + ( n p ) / a 2 ) = x 3 + [ ( n p / a 2 ) − 2 a ] x 2 + ( a 2 − 2 n p / a ) x + n p . F r o m ( n p / a 2 ) − 2 a = 0 , w e g e t p = 2 a 3 / n , a n d f r o m a 2 − 2 n p / a = p , − 3 a 2 = 2 a 3 / n , n = 2 a / − 3 , s m a l l e s t n = 2 , f o r a = − 3. S o p = − 27. f ( x ) = ( x + 3 ) 2 ( x − 6 ) = x 3 − 27 x − 54 = 0 , p + q = − 27 − 54 = − 81 In order to understand your work, I have copied your work as above. f(x) =~~ (x - a)^2(x+(np)/a2)\\ =~~ x^3 + [(np/a^2) - 2a]x^2 + (a^2 - 2np/a) x +np.\\ From~ (np/a^2) - 2a = 0,~ we ~get~~~ p=2a^3/n,~ and~ from ~~~a^2 - 2np/a = p,\\ - 3a^2 = 2a^3/n,~~~~ n=2a/-3,~ smallest~ n=2,~ for~ a=-3.~~So ~~p=-27.\\ f(x)=(x+3)2(x - 6) = x3 - 27x - 54 =0,~~ p+q= -27 - 54 = -81\\ \text{In order to understand your work, I have copied your work as above.}

Niranjan Khanderia - 5 years, 6 months ago

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