Given that x 3 − 8 x 2 + a x + b = 0 has integer roots. If a 2 − 3 6 a + 3 2 3 = 0 , find a + b .
The answer is 7.
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3 solutions
"because maybe sum of numbers being 0" is questionable. Which is the sum you are talking about?
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oh,sorry I could not type this Polynomial,when a+b+c+d=0 one of roots is 1!So 1+(-8)+17+(-10)=0, 1+(-8)+19+(-12)=0 so one of roots is 1! also polynomial (2):like axx+bx+c=0 ,when a+b+c=0 one of roots is 1. I typed b=(- with shift,10) but didnt show this so i edit this and I type b=-10,-12.
This is some lucky guesswork. You failed to show that a + b = 7 only.
We first solve for a : a 2 − 3 6 a + 3 2 3 = a 2 − 1 7 a − 1 9 a + 3 2 3 = a ( a − 1 7 ) − 1 9 ( a − 1 7 ) = ( a − 1 9 ) ( a − 1 7 ) = 0 . Thus, a = 1 7 , 1 9 .
We want to find the relationship between a and b :
x 3 − 8 x 2 + a x + b = 0 ⇒ x ( x 2 − 8 x + a ) = − b ⇒ x 2 − 8 x + a = − x b ⇒ ( x − 4 ) 2 + a − 1 6 = − x b
For x = 4 , the equation simplifies to a − 1 6 = − 4 b .
If
a
=
1
7
, then
b
=
−
4
.
If
a
=
1
9
, then
b
=
−
1
2
.
However, for ( a , b ) = ( 1 7 , − 4 ) , it's easy to verify that the equation x 3 − 8 x 2 + 1 7 x − 4 = 0 can't have ALL integer roots by Rational Root Theorem .
Hence, a = 1 9 , b = − 1 2 only, thus a + b = 7 only.
I am afraid a cubic equation like this may or may not be having 4 as a root, so your assuming that a and b retain the relationship as inferred by putting x = 4 is not correct. You have reached the correct answer because x = 4 happened to be one of the roots under some condition.
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No. I can put in any value other than x = 0 to find the relationship between a and b. I can put in x = 1 (which is simpler, come to think of it) and it will also yield the answer of a=19, b = -12.
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But how come am i getting 3 solutions, 5, 2 and 1 when i take a=17 and b =-10?
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@A Former Brilliant Member – What seems to be the problem? You get a + b = 7 as well? Or are you concerned that I get a = 1 7 , b = − 4 and you get a = 1 7 , b = − 1 0 ? Then think of it this way: since you substitute a = 1 7 already, then you can substitute any values of x into the first equation to find the value of b .
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@Pi Han Goh – Oh naah! That's not the problem! You wrote in your solution, the word 'only' which kinda confused me! :-P
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@A Former Brilliant Member – It's only true when x = 4 . If you substitute x as something else, you will possibly get another set of ( a , b ) .
x = 1 is also one of those that happen to be a root of this equation. That is why the relation between a and b is retained. It is not just any value, as you claim.
Just substitute x for 1 at the first equation.
Moderator note:
Your solution makes the implicit assumption that variables ( a , b , x ) exist to satisfy the given conditions. Thus, we should verify that the assumption is valid, to make the solution complete.
(a-17)(a-19) So a=17 , a=19 So I was lucky,I choosed b=-10 , b=-12 because maybe sum of numbers being :0,then one of root being :1.So that is true & a+b=7.
so another roots is x=2 , x=5 for a=17&b=-10 and for a=19&b=-12 another roots is x=3 , x=4 hahahahaha!!