Polynomials with some unknown
Find all integers a such that the cubic equation x 3 − 3 x + a , has only integer roots.
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1 solution
if a=-2, then discriminant comes out to be 17 which is not a square... so tell me if u put a=-2, how can u get integer roots? @Shivam Jadhav
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This is a cubic not a quadratic.
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Well first of all it is not stated that the integer roots are same or different
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@Aayush Patni – Well,I can't add this solution in the solutions' section as my dad saw the solution while I was still trying.:P.So here it is: Let us assume that : x 3 − 3 x + a = ( x + b ) ( x 2 + c x + b a ) Simplifying the R.H.S and comparing co-efficients gives : b = ( − c ) . . . . . . ( 1 ) and − 3 = b c + b a . . . . . . ( 2 ) . The roots become (-b) and the roots of the quadratic equation in the R.H.S . The roots of the quadratic equation are : 2 − c ± c 2 − b 4 a = 2 − c ± b c 2 b − 4 a Now,in place of b c 2 , we can put b 3 , ( from 1 ) and,in place of 4a,we can put 4 b 3 − 1 2 b ( from (1) and (2) ) putting this in the above expression we get, 2 − c ± 1 2 − 3 b 2 . Now, 1 2 − 3 b 2 ≥ 0 ⟹ 4 ≥ b 2 . This expression gives 5 values of b,namely,-2,-1,0,1,2 . Now just do hit and trial. .P.S:It is 12 in the night and I have school tomorrow :P:P
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@Adarsh Kumar – @Calvin Lin @Shivam Jadhav @Archit Boobna
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@Adarsh Kumar – For the polynomial y = x 2 − 2 x + 1 , we say that it has 1 root (since it repeats) which is 1.
So in context to the above question, if I get the roots 2,1 and 1, shouldn't I say the polynomial has 2 roots . Since 1 is repeated, it should be counted as 1 root.
So technically, the answer should be none of these, since we hget 2 integer roots.
Please do correct me if I am wrong. Adarsh Kumar Calvin Lin
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@Pankaj Joshi – even i did the same....
@Pankaj Joshi – We say that it has 2 roots, and that these are repeated roots. There is only 1 distinct roots.
In general, a polynomial of degree n will have n (complex) roots.
@Aayush Patni – Yes! That's Why I got it wrong
well i could not add a solution......as i got the wrong answer first...
for solving this ....first we differentiate the function..... we get 3(x-1) thus we conclude that the function has slope=0 at 1 and -1 and for the cubic to have 3 roots which are integer...it should intersect x-axis 3 times...or touch x axis one time and intersect it one time...... if we look at the situation graphically.....then....for the roots to be integral...one of the roots should either be 0 or 1 or -1.........because since the slope it zero at -1 and 1...the graph certainly intersects the x-axis somewhere between -1 and 1...or at 1 or at -1............ and in case of 1 and -1.............if the function have them as its roots then certainly ...the function would have only two distinct roots as....the slope at -1 and 1 is zero...... so firstly checking for zero....putting x=0.....and equating the expression with zero...we get a=0 ....but the other roots obtained are not integral......so zero rejected......
secondly putting x=-1 equating the expression with zero..........we get a=-2 now we know that the product of roots is 2...and the function has two distinct roots....where -1 is the repeated root....we get the other root to be 2..by dividing 2 by (-1*-1) hence a=-2 accepted... similarly doing for x=1 we get a=2 which satisfies the conditions
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@Calvin Lin @Shivam Jadhav @Archit Boobna @ Adarsh Kumar
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@Yash Sharma I totally agree with you and even I got that answer. But it should have been mentioned that same roots are also considered as 2 roots.
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@Archit Boobna – i think i did..............i wrote the same as repeated root.............
Using vieta's theorem We get 0 ≤ m 2 , n 2 , l 2 ≤ 6 Hence m=2 , n=-1,l=-1 And m= -2,n=1,l=1 m,n,l are roots of given equation.