Conclude and Prove

Algebra Level 3

Given that p p and q q are both prime, which of the following answer choices is true about the equation p x 2 q x + q = 0 ? px^{ 2 }-qx+q=0?

Has rational solutions Has exactly one solution Has no rational solutions Insufficient data

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10 solutions

For there to be rational solutions we would require that the discriminant

q 2 4 p q = q ( q 4 p ) q^{2} - 4pq = q(q - 4p)

is a perfect square. For this to be the case, since q q is prime, we would require that q 4 p = q n 2 q ( 1 n 2 ) = 4 p q - 4p = qn^{2} \Longrightarrow q(1 - n^{2}) = 4p for some non-negative integer n n .

Now if n = 0 n = 0 we would have q = 4 p q = 4p . However, since both p p and q q are prime we can't have p q p | q , and thus n n cannot be 0 0 .

If n 1 n \ge 1 then q ( 1 n 2 ) 0 q(1 - n^{2}) \le 0 , while 4 p > 0 4p \gt 0 . Thus n n cannot be a positive integer.

We can then conclude that the discriminant cannot be a perfect square, and thus the given quadratic has no rational solutions.

@brian charlesworth sir, you can just say that q × ( 1 n 2 ) q\times (1-n^{2}) will be negative for all n > 1 n>1 .

Adarsh Kumar - 6 years, 6 months ago

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Yes, but then you would still have to deal with the case n = 1 n = 1 , which is why I combined the cases in saying that for n 1 n \ge 1 we have q ( 1 n 2 ) 0 q(1 - n^{2}) \le 0 .

Brian Charlesworth - 6 years, 6 months ago

sir , can you help with this:There exists a quadratic polynomial a x^2 + b x + c=0 ,a,b,c belong to odd positive integers.Prove that there are no rational solutions.

Raven Herd - 5 years, 11 months ago

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For there to be rational solutions we would require that the discriminant be a perfect square. So suppose that b 2 4 a c = n 2 b^{2} - 4ac = n^{2} for some positive integer n . n.

Note first that, with a , b , c a,b,c all odd, we would have b 2 4 a c b^{2} - 4ac being odd, and thus n n would be odd as well. Now rewrite the equation as

b 2 n 2 = 4 a c ( b n ) ( b + n ) = 4 a c . b^{2} - n^{2} = 4ac \Longrightarrow (b - n)(b + n) = 4ac.

Now since a , c a,c are both odd, 4 a c 4ac has exactly two factors of 2. 2. For the left-hand side, since b , n b,n are both odd, both ( b n ) (b - n) and ( b + n ) (b + n) are even and differ by 2 n . 2n. Now since n n is odd, we know that 2 n 2 ( m o d 4 ) . 2n \equiv 2 \pmod{4}. Thus one or the other of ( b n ) (b - n) or ( b + n ) (b + n) must be divisible by 4 , 4, which implies that ( b n ) ( b + n ) (b - n)(b + n) has at least three factors of 2. 2. Since 4 a c 4ac has exactly two factors of 2 , 2, we can conclude by contradiction that there exists no positive integer n n such that b 2 4 a c = n 2 b^{2} - 4ac = n^{2} where a , b , c a,b,c are all odd, and thus a x 2 + b x + c = 0 ax^{2} + bx + c = 0 has no rational roots when a , b , c a,b,c are all odd, positive integers.

Brian Charlesworth - 5 years, 11 months ago

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I don't get why should one of (b-n),(b+n) be divisible by 4?

Raven Herd - 5 years, 11 months ago

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@Raven Herd First, they are both even, so are both divisible by 2 , 2, and can either be 0 0 or 2 2 mod 4. 4. So suppose ( b n ) 2 ( m o d 4 ) . (b - n) \equiv 2 \pmod{4}. Then since 2 n 2 ( m o d 4 ) , 2n \equiv 2 \pmod{4}, we have that

( b + n ) = ( ( b n ) + 2 n ) ( 2 + 2 ) ( m o d 4 ) 0 ( m o d 4 ) . (b + n) = ((b - n) + 2n) \equiv (2 + 2) \pmod{4} \equiv 0 \pmod{4}.

Similarly, if ( b + n ) 2 ( m o d 4 ) (b + n) \equiv 2 \pmod{4} then

( b n ) = ( ( b + n ) 2 n ) ( 2 2 ) ( m o d 4 ) 0 ( m o d 4 ) . (b - n) = ((b + n) - 2n) \equiv (2 - 2) \pmod{4} \equiv 0 \pmod{4}.

So both b n b - n and b + n b + n are even and precisely one of them must be divisible by 4 , 4, implying that their product is divisible by 8 , 8, i.e., their product must have at least three factors of 2. 2.

Brian Charlesworth - 5 years, 11 months ago

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@Brian Charlesworth most brilliantly done as always! Sir, when I started doing this problem I reached the fact that both (b-n),(b+n) differ by 2 ,but after that I couldn't go further.My maths teacher tells me that I must generate the problem solving ability in myself .Is it that easy? Just like you wake up one day and find all the intelligence into yourself .He also says that there are infinite dimensions to a problem , if you do it yourself no matter how small it is but it will supply you with something new.Those were some nice words which I truly appreciate.When there is a really nice problem ,most of the time I end up reading the solution which I somehow manage to understand .Please give your opinion on this.

Raven Herd - 5 years, 11 months ago

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@Raven Herd Your teacher has given you some good advice. Developing ones' problem solving ability is a gradual process; it comes both from continuous practice and from reading and comprehending the elegant solutions written by others. After solving a problem one way, it's always a good exercise to find as many other ways to solve the same problem as you can, (which I think is what your teacher meant by "infinite dimensions"). Also, as you have done here, after solving a problem you can pose yourself a different (and more difficult) version of the question and see if you can solve it. One good question always leads to another.

If I can't solve a problem at first, I'll set it aside and come back to it later. If I still can't solve it later, then I'll set it aside again for even later. I'm very stubborn and tend not to reveal solutions to questions even after many attempts, but I don't think that's the best approach for learning; it's probably better to reveal a solution and learn from someone else, just as you do, than it is to be too stubborn like me. :)

Brian Charlesworth - 5 years, 11 months ago

@Brian Charlesworth Brilliant solution! So well thought out!

Krish Shah - 1 year, 1 month ago

that is just awesome!! \text{that is just awesome!!}

Adarsh Kumar - 6 years, 6 months ago

Just take the following equation and you will find that it do have rational solution. 3x^2 - 23x + 23 = 0. you will find that 23^2-4 3 23 > 0.

Chitranshu Vashishth - 6 years, 6 months ago

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In this case, the discriminant is 2 3 2 4 3 23 = 253 23^{2} - 4*3*23 = 253 , which is greater than zero but is not a perfect square. The roots of the equation are

x = 23 ± 253 6 x = \dfrac{23 \pm \sqrt{253}}{6} ,

which are not rational.

Brian Charlesworth - 6 years, 6 months ago

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what about 2x^2 - 17x + 17 ?

Abhinav Kumar Santoshi - 5 years, 8 months ago

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@Abhinav Kumar Santoshi Then the roots would be x = 17 ± 3 17 4 , x = \dfrac{17 \pm 3\sqrt{17}}{4}, which are not rational.

Brian Charlesworth - 5 years, 8 months ago

Sorry ,not legible 4 me

Momina Ghanchi - 6 years, 5 months ago

though i did it correctly, i always check for your solution, and as always your explanation is best. i am happy that i was close to you.thank you sir, for you are here to explain these math problems.

manish kumar singh - 5 years, 7 months ago

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My pleasure, Manish. Thanks for the compliment. :)

Brian Charlesworth - 5 years, 7 months ago

It is clear that there could be imaginary solutions (simple case p and q are the same prime numbers). You mean to tell me that a complex number is irrational?

Federico Martin - 5 years, 4 months ago

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The terms rational and irrational only apply to real numbers. A complex number a + b i a + bi can have rational and irrational parts a , b a,b but we wouldn't generally refer to the complex number itself as being either rational or irrational. I've shown that any real solution will not be rational, and from my previous comment any complex solution will by definition neither be rational nor irrational, but just simply complex.

Now we could have a complex solution that has rational parts, however. For example, with p = 5 , q = 2 p = 5, q = 2 we have the equation 5 x 2 2 x + 2 = 0 5x^{2} - 2x + 2 = 0 , which has solutions

x = 2 ± 4 40 10 = 2 ± 6 i 10 = 1 5 ± 3 5 i x = \dfrac{2 \pm \sqrt{4 - 40}}{10} = \dfrac{2 \pm 6i}{10} = \dfrac{1}{5} \pm \dfrac{3}{5}i .

Brian Charlesworth - 5 years, 4 months ago

This might sound stupid but why did we consider n to be a non-negative integer? I know it wouldn't effect the answer but still

Vandit Kumar - 3 years, 3 months ago

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I was thinking the same thing.

Krish Shah - 1 year, 1 month ago

Let's have 2 roots A and B. AB=q since q is prime A has to be 1 or B has to be 1. A+B=-q; q has to be either 2 or an odd prime. The only possibility for an odd prime are if roots are 2 and 1. This gives equation (x-2)(×-1)=×2-3×+2. Which has rational roots. The roots can't be 1 and 1 because 1 isn't a prime. I'm confused as to why this is wrong

Ashish Sacheti - 5 years, 11 months ago

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The polynomial x 2 3 x + 2 x^{2} - 3x + 2 is not in the desired form p x 2 q x + q px^{2} - qx + q for primes p , q . p,q. Also, regarding your first sentence, while it is true that A B = q AB = q this does not mean that A A or B B has to be 1 , 1, since we could be looking for rational roots, (and not necessarily integer roots).

Brian Charlesworth - 5 years, 11 months ago

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LOL sorry i had a brain fart and read equation wrong. This makes sense thanks

Ashish Sacheti - 5 years, 11 months ago

AB should be q/p. Don't you think so, sir?

Himanshu Agrawal - 5 years, 10 months ago

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@Himanshu Agrawal Yes, that's true, and also A + B = q p . A + B = \frac{q}{p}. So unless q = p q = p both A B AB and A + B A + B will be non-integral. (If p = q p = q then we would, after simplification, have the equation x 2 x + 1 = 0 , x^{2} - x + 1 = 0, which has the complex solutions 1 ± i 3 2 . \frac{1 \pm i\sqrt{3}}{2}. ) Regardless, for an equation in the form given in the question, the roots are not rational.

Brian Charlesworth - 5 years, 10 months ago

I have a similar solution. For q 2 4 p q q^2-4pq to be a perfect square, p p must be zero which contradicts the given condition. Thus, no answer.

Roman Frago - 5 years, 11 months ago
Al Imran
Dec 11, 2014

Here, determinant of the equation is √(q^2-4pq)=√q*√(q-4p). Since, q and p are prime, therefore, it is irrational. Therefore, roots are irrational.

Best approach man...........the easiest one......

MAK JUNIOR - 6 years, 5 months ago

I'm not satisfied with this way of proving the irrationality as q & p are prime, doesn't mean that sqrt(q - 4p) sqrt(q) will always not be a perfect square… I'm not against answer but against method. After applying this method, there will still doubt whether sqrt(q - 4p) sqrt(q) will be integer or not? And if you still sure that you are right then plz explain your view in detail.

Himanshu Agrawal - 5 years, 10 months ago

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well it's obvious that √q is irrational because q is prime thus the product is irrational and so are the roots

Abdessabour M - 5 years, 6 months ago

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Yes s q r t ( q ) sqrt(q) is irrational but so can s q r t ( q 4 p ) sqrt(q-4p) . For example, s q r t ( 3 ) sqrt(3) and s q r t ( 27 ) sqrt(27) are irrational but s q r t ( 81 ) sqrt(81) is not irrational.

Krish Shah - 1 year, 1 month ago
Tom Schulte
Dec 18, 2014

By the Rational Zeros Theorem, for this particular quadratic the only possible rational solutions are plus and minus q / p . Regardless of the sign, plugging in either case for x simplifies to q = 0 which is a contradiction.

Pablo Padilla
Nov 26, 2015

By the theorem of rational roots, the only rational solutions for this equation could be q p \frac{q}{p} and q p -\frac{q}{p} , since p p and q q are prime.

Substituting, we find that the condition for any of them to satisfy the equation is that q = 0 q=0 , but that is a contradiction since 0 is not prime.

I did the same way. It is much more faster and in my opinion, elegant.

Anupam Nayak - 5 years, 5 months ago
Sitta Ramona
Dec 10, 2014

X = (q± √(q^2-4pq))/2p, because q and p are prime, q^2 – 4pq cannot be a perfect square, so the given quadratic has no rational solutions.

Andrea Palma
Mar 9, 2016

Ruffini's criterion for a polinomyal with integer coefficients

a n x n + a n 1 n 1 + + a 1 x + a 0 a_nx^n + a_{n-1}^{n-1}+ \cdots + a_1x + a_0

to have rational roots is that these roots have the following form ± a b \pm \dfrac{a}{b} with a a a divisor of a 0 a_0 and b b a divisor of a n a_n .

So we have to prove all rationals ± a b \pm \dfrac{a}{b} with a = 1 , q a = 1, q and b = 1 , p b = 1, p . We can exclude the negative numbers couse the result would be greater than 0 0 . By mere proof we find that no such number is a root, so the polinomyal has no rational roots.

Edwin Gray
May 8, 2018

If p = q, then qx^2 - qx - q = 0. Since q .ne. 0, dividing by q, x^2 - x - 1 =0, having no rational roots. If(p,q) =1, then x|q and q|x, so q =x. Then pq^2 - q^2 - q =0 which implies q^2|q, which is impossible, so no rational root. Ed Gray

Dervin Bremont
Jul 13, 2016

This is not a good solution but I use logic : - A Insufficient data (Is not a solution). - B Has no rational solutions - C Has rational solutions - D Has exactly one solution

So the solutions could be , B, C, or D. - Assume that D is the solution, then you have that C is also other solution, then there is two solutions, but the solution is only one, this imply that D is not a solution.

  • The solutions could be : B, C. I just guess B and It works. (I do that because I don´t know enough math for solve these problem.)
Andrew Lovelock
May 27, 2016

By Eisenstein's criterion the polynomial is irreducible in Q ( x ) \mathbb{Q}(x) , so has no rational roots. (Okay, when p and q are equal we have to discount that case but it's not hard).

Short and simple! Thanks for sharing your approach, Andrew.

Pranshu Gaba - 5 years ago
Anthony Zamberlan
Jan 22, 2016

Clear by Eisenstein's using the prime q.

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