Roots of a cubic function
Let a , b , c , d , e and f be constant real numbers such that the roots (for x ) in x 3 + a x 2 + b x − 1 0 5 = 0 are each two more than the roots (for x ) in x 3 − d x 2 + e x − f = 0 . Determine the value of 4 d + 2 e + f .
The answer is 97.
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4 solutions
Let r, s, t be the roots of the first equation and u, v, w for the second.
105 = rst = 7 * 3 * 5
u = 7-2=5
v = 5-2 =3
w =3-2=1
f = uvw= 5 * 3 * 1=15
e = uv * vw * uw = 5*3 + 3 * 1 +1 * 5=23
d = u+ v+ w = 1+3+5= 9
4d + 2e + = 97
Typo last line, f +4d+2e=97.
L e t p , q , r b e t h e r o o t s o f s e c o n d e q u a t i o n . ∴ p + 2 , q + 2 , r + 2 a r e t h e r o o t s o f f i r s t e q u a t i o n . ⟹ f r o m V i e t a ′ s f o r m u l a : − S e c o n d e q u a t i o n : − p + q + r = d , p q + q r + r p = e , p q r = f . F i r s t e q u a t i o n : − ( p + 2 ) ( q + 2 ) ( r + 2 ) = 1 0 5 B u t ( p + 2 ) ( q + 2 ) ( r + 2 ) = p q r + 2 ( p q + q r + r p ) + 4 ( p + q + r ) + 8 = 1 0 5 ⟹ 1 0 5 = f + 2 e + 4 d + 8 = 1 0 5 . S o 4 d + 2 e + f = 1 0 5 − 8 = 9 7 .
Let the roots of x 3 + a x 2 + b x − 1 0 5 = 0 be p , q , r . From Vieta's formula, we have
p + q + r = − a p q + p r + q r = b p q r = 1 0 5
Now, we know that the roots of x 3 − d x 2 + e x − f = 0 are p − 2 , q − 2 , r − 2 . Again, from Vieta's formula, we have:
( p − 2 ) + ( q − 2 ) + ( r − 2 ) = d p + q + r − 6 = d d = − a − 6
( p − 2 ) ( q − 2 ) + ( p − 2 ) ( r − 2 ) + ( q − 2 ) ( r − 2 ) = e p q − 2 p − 2 q + 4 + p r − 2 p − 2 r + 4 + q r − 2 q − 2 r + 4 = e p q + p r + q r − 4 ( p + q + r ) + 1 2 = e e = b − 4 ( − a ) + 1 2 = b + 4 a + 1 2
( p − 2 ) ( q − 2 ) ( r − 2 ) = f ( p q − 2 p − 2 q + 4 ) ( r − 2 ) = f p q r − 2 p r − 2 q r + 4 r − 2 p q + 4 p + 4 q − 8 = f p q r − 2 ( p q + p r + q r ) + 4 ( p + q + r ) − 8 = f f = 1 0 5 − 2 b + 4 ( − a ) − 8 = 9 7 − 2 b − 4 a
4 d + 2 e + f = 4 ( − a − 6 ) + 2 ( b + 4 a + 1 2 ) + 9 7 − 2 b − 4 a = − 4 a − 2 4 + 2 b + 8 a + 2 4 + 9 7 − 2 b − 4 a = 9 7
Let ( α , β , γ ) be the roots of f ( x ) = x 3 + a x 2 + b x − 1 0 5 ( We let f ( x ) = x 3 + a x 2 + b x − 1 0 5 ). Therefore, ( α − 2 , β − 2 , γ − 2 ) are the roots of x 3 − d x 2 + e x − f .
But then, ( α − 2 , β − 2 , γ − 2 ) are the roots of f ( x + 2 ) , since f ( ( α − 2 ) + 2 ) = f ( α ) = 0 .
Therefore, ( α − 2 , β − 2 , γ − 2 ) are the roots of f ( x + 2 ) = ( x + 2 ) 3 + a ( x + 2 ) 2 + b ( x + 2 ) − 1 0 5 = x 3 + ( a + 6 ) x 2 + ( 1 2 + 4 a + b ) x + ( 4 a + 2 b − 9 7 ) .
This implies that x 3 + ( a + 6 ) x 2 + ( 1 2 + 4 a + b ) x + ( 4 a + 2 b − 9 7 ) = x 3 − d x 2 + e x − f . Comparing coefficients, we get that d = − a − 6 , e = 1 2 + 4 a + b and f = 9 7 − 4 a − 2 b .
Therefore, our required answer is 4 d + 2 e + f = 4 ( − a − 6 ) + 2 ( 1 2 + 4 a + b ) + ( 9 7 − 4 a − 2 b ) = 9 7 + 2 4 − 2 4 + 8 a − 8 a + 2 b − 2 b = 9 7 .