A cubic, how depressing
The cubic polynomial
x 3 − 9 x 2 + 2 0 x + 2
can be expressed in terms of y as
y 3 + c y + d ,
where c and d are real numbers and y has a linear relationship with x .
What is the value of c + d ?
The answer is 1.
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4 solutions
Here I propose a very simple solution Consider the cubic function: f ( x ) = x 3 − 9 x 2 + 2 0 x + 2
Now, transforming the cubic function to the depressed form is by substituting x = y − b / 3 a
which in this case evaluates to x = y − ( − 9 ) / 3 ( 1 ) = y + 3
Hence,
f ( y + 3 ) = y 3 + c y + d
Sub y = 1
f ( 4 ) = 1 + c + d
Evaluating f ( 4 ) gets 2
So
c + d = 2 − 1 = 1
Surprised you didn't get any upvotes. Good job
We can write the equation as,
x
3
−
9
x
2
+
2
0
x
+
2
=
0
.
x
3
−
9
x
2
+
2
7
x
−
2
7
−
2
7
x
+
2
7
+
2
0
x
+
2
=
0
.
(
x
−
3
)
3
−
7
x
+
2
9
=
0
.
(
x
−
3
)
3
−
7
x
+
2
1
+
8
=
0
.
(
x
−
3
)
3
−
7
(
x
−
3
)
+
8
=
0
.
Comparing this equation with
y
3
+
c
y
+
d
=
0
we get,
c
=
−
7
,
d
=
8
and the linear relation as
y
=
x
−
3
.
Therefore,
c
+
d
=
−
7
+
8
=
1
.
put /$ x=y+9/3 /$ /newline $ (y+3)^3-9(y+3)^2+20(y+3)+2= $ /newline $ y^3+9y^2+27y+27-9y^2-54y-81+20y+60+2= $ /newline $ y^3-7y+8 $
Relevant wiki: Cardano's Method
Assuming you don't know the identity that depresses the cubic , let x = y − p . Then the cubic is:
x 3 − 9 x 2 + 2 0 x + 2 ( y − p ) 3 − 9 ( y − p ) 2 + 2 0 ( y − p ) + 2 y 3 − 3 p y 2 + 3 p 2 y − p 3 − 9 ( y 2 − 2 p y + p 2 ) + 2 0 y − 2 0 p + 2 y 3 + ( − 3 p − 9 ) y 2 + ( 3 p 2 + 1 8 p + 2 0 ) y + ( − p 3 − 9 p 2 − 2 0 p + 2 ) .
The 2nd degree term should have a 0 coefficient to fit the required form. Thus,
− 3 p − 9 p = 0 = − 3 .
Substituting this value of p back into the expression gives:
y 3 − 7 y + 8 .
Thus, c + d = 1 .