Two cubic equations
If a 3 − 3 a 2 + 5 a − 1 7 = 0 and b 3 − 3 b 2 + 5 b + 1 1 = 0 are such that ( a + b ) is a real number, then find the value of ( a + b ) .
The answer is 2.
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2 solutions
can you please explain Cardano's method?
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You can read about it here .
@avn bha Thats a powerful but quite calculative method for solving cubic equations.
@Chew-Seong Cheong I think Cardano's method need not be mentioned. It was a mere transformation, though inspired by Cardano's method.
@Sandeep Bhardwaj Nice question. :)
I have posted a wiki on Cardano's Formula . Hope it is useful to you.
This is not Cardano's method. What you did was you just depressed the cubic equations, that's all. Plus, you need to explain why x 2 + y 2 − x y + 2 = 0 cannot be a solution. So unfortunately this solution does not work.
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The solution works . We can explain that x 2 + y 2 − x y + 2 = 0 will not have any solution by considering it a quadratic in x or y and showing that the discriminant is less than zero which will mean that it does not have real roots.
x 2 − x y + ( y 2 + 2 ) = 0
The discriminant will be ,
D = y 2 − 4 ( 2 + y 2 )
D = − 8 − 3 y 2
Which will be always negative . Hence proved
But how did it strike you to do it this way?
what i thought was something different and i don't know whether its correct or not. Please correct me if i am wrong
let the roots of 1st equation be x,y,z ,and that of 2nd equation be m,n,q. by vietas we have x+y+z =3=m+n+q adding=x+y+z+m+n+q=6
now since x+m=a+b=y+n=z+q there fore 3(a+b)=6
(a+b)=2
How can you say that x + m = a + b = y + n = z + q ?
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I didn't say that .if you take any of the 3 pairs that is even if you take x+n=a+b=y+m that would be true.
Using Cardano's method for cubic equations, substituting:
{ a = x + 1 b = y + 1 ⇒ { a 3 − 3 a 2 + 5 a − 1 7 = 0 b 3 − 3 b 2 + 5 b + 1 1 = 0 ⇒ { x 3 + 2 x = 1 4 y 3 + 2 y = − 1 4
Adding the last two equations together, we have:
x 3 + y 3 + 2 ( x + y ) = 0 ⇒ ( x + y ) ( x 2 + y 2 − x y ) + 2 ( x + y ) = 0
⇒ ( x + y ) ( x 2 + y 2 − x y + 2 ) = 0 ⇒ x + y = 0
⇒ a + b = x + 1 + y + 1 = x + y + 2 = 0 + 2 = 2