Where to Start?

Algebra Level 5

For a fixed pair of real numbers ( b , c ) (b, c) with b < 0 b < 0 , find the number of possible values of a a that will satisfy the following conditions:

The equation x 4 + a x 3 + b x 2 + c x + d = 0 { x }^{ 4 }+{ ax }^{ 3 }+{ bx }^{ 2 }+cx+d=0 with a , b , c , d ∈ R a,b,c,d \in R has roots x 1 , x 2 , x 3 , x 4 { x }_{ 1 },{ x }_{ 2 },{ x }_{ 3 },{ x }_{ 4 } such that x 1 + x 2 = x 3 + x 4 { x }_{ 1 }+{ x }_{ 2 }={ x }_{ 3 }+{ x }_{ 4 } .


The answer is 1.

View wiki
Deepanshu Gupta by Deepanshu Gupta , 25, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Deepanshu Gupta
Dec 3, 2014

Using Theory Of Equation :

− a = x 1 + x 2 + x 3 + x 4 x 1 + x 2 = x 3 + x 4 = − a 2 . . . ( 1 ) b = ∑ x 1 x 2 b = x 3 ( x 1 + x 2 ) + x 4 ( x 1 + x 2 ) + x 1 x 2 + x 3 x 4 x 1 x 2 + x 3 x 4 = b − a 2 4 . . . ( 2 ) − c = ∑ x 1 x 2 x 3 − c = x 1 x 2 ( x 3 + x 4 ) + x 3 x 4 ( x 1 + x 2 ) c = a 8 ( 4 b − a 2 ) . . . . ( 3 ) -a\quad =\quad { x }_{ 1 }+\quad { x }_{ 2 }\quad +\quad { x }_{ 3 }\quad +{ x }_{ 4 }\\ \\ { x }_{ 1 }+\quad { x }_{ 2 }\quad =\quad \quad { x }_{ 3 }\quad +{ x }_{ 4 }\quad =\quad \cfrac { -a }{ 2 } \quad \quad .\quad .\quad .(1)\\ \\ b\quad =\quad \sum { { x }_{ 1 }{ x }_{ 2 }\quad } \\ \quad \quad \\ b\quad =\quad { x }_{ 3 }({ x }_{ 1 }+\quad { x }_{ 2 })\quad +\quad { x }_{ 4 }({ x }_{ 1 }+\quad { x }_{ 2 })\quad +\quad { x }_{ 1 }{ x }_{ 2 }\quad +\quad { x }_{ 3 }{ x }_{ 4 }\\ \\ \\ \quad { x }_{ 1 }{ x }_{ 2 }\quad +\quad { x }_{ 3 }{ x }_{ 4 }\quad =\quad b\quad -\quad \cfrac { { a }^{ 2 } }{ 4 } \quad \quad \quad \quad \quad .\quad .\quad .(2)\\ \\ -c\quad =\quad \sum { { x }_{ 1 }{ x }_{ 2 }{ x }_{ 3 }\quad } \\ \\ -c\quad =\quad { x }_{ 1 }{ x }_{ 2 }({ x }_{ 3 }\quad +{ x }_{ 4 })\quad +\quad { x }_{ 3 }{ x }_{ 4 }({ x }_{ 1 }+\quad { x }_{ 2 })\quad \\ \\ \quad c\quad =\quad \cfrac { a }{ 8 } (4b\quad -\quad { a }^{ 2 })\quad \quad .\quad .\quad .\quad .\quad (3) .

Then Let the 3rd equation as f(a)

f ( a ) = a 3 − 4 a b + 8 c = 0 f ′ ( a ) = 3 a 2 − 4 b ∵ b < 0 f ′ ( a ) > 0 f\left( a \right) \quad =\quad { a }^{ 3 }\quad -\quad 4ab\quad +\quad 8c\quad =\quad 0\\ \\ f'(a)\quad =\quad 3{ a }^{ 2 }\quad -\quad 4b\\ \\ \because \quad b\quad <\quad 0\quad \\ \\ f'(a)\quad >\quad 0\quad .

So this implies f(a) has one root ( Since Monotonic) ( Draw it's Graph )

So For Given Fixed Pair ( b , c ) ' a ' has only one value

11 Helpful 0 Interesting 0 Brilliant 0 Confused

In your case 1, for each pair of values of ( b , c ) (b,c) , there is 1 corresponding real value of a a . Note that if we fix b b and let c c vary, then this root clearly changes. Hence, there are infinitely many possible real values of a a .

In fact, what you have shown, is that if ( b , c ) (b, c) are fixed values with b < 0 b < 0 , there there is a unique value of a a such that the roots can pair up to give x 1 + x 2 = x 3 + x 4 x_1 + x_2 = x_3 + x_4 . This is an interesting result.

Calvin Lin Staff - 6 years, 6 months ago

Log in to reply

@Calvin Lin Sir I edited Question Accordingly ! Can You Please reviewed it !

Thanks !

Deepanshu Gupta - 6 years, 6 months ago

Thanks. I have updated the answer to 1. Please update your solution accordingly.

The previous statement of the problem had an answer of infinity. Those who answered 2 have been marked incorrect.

Calvin Lin Staff - 6 years, 6 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...