# Finding three X values for y=ax^3+bx^2+cx+d

• March 24th 2010, 12:43 AM
Neverquit
Finding three X values for y=ax^3+bx^2+cx+d
Hello,
I am trying to solve for the x's for y=ax^3+bx^2+cx+d.
It cuts the y-axis at (0,30). The gradient at this point is -1.
The gradient where x=1, is -17 and the f ``(x) at this point is -10.

I s.t the coordinates given; set f `(x) =-1 and
f ``(x)=-10.

I think d=30 and c=-1.
So I need to solve for a,b.

I didn't think solving for x would be this tricky? What have I missed?

• March 24th 2010, 12:54 AM
Pulock2009
y=0 to solve
Quote:

Originally Posted by Neverquit
Hello,
I am trying to solve for the x's for y=ax^3+bx^2+cx+d.
It cuts the y-axis at (0,30). The gradient at this point is -1.
The gradient where x=1, is -17 and the f ``(x) at this point is -10.

I s.t the coordinates given; set f `(x) =-1 and
f ``(x)=-10.

I think d=30 and c=-1.
So I need to solve for a,b.

I didn't think solving for x would be this tricky? What have I missed?

if you are trying to solve for x simply solve for y=0. you will get three values of x in terms of a,b,c and d. refer to theory of equations . solve the following equations obtained by the theory:
x1+x2+x3=-b/a------------(1)
x1.x2+x2.x3+x3.x1=c/a----------(2)
x1.x2.x3=-d/a------------------(3)
where x1,x2,x3 are the 3 roots.also given that d=30 from the given point.
• March 24th 2010, 07:05 AM
Neverquit
Hi,
I am still unsure about how to apply the theory of equations to this problem. Any further information would be appreciated.
• March 24th 2010, 07:25 AM
skeeter
Quote:

Originally Posted by Neverquit
Hello,
I am trying to solve for the x's for y=ax^3+bx^2+cx+d.
It cuts the y-axis at (0,30). The gradient at this point is -1.
The gradient where x=1, is -17 and the f ``(x) at this point is -10.

I s.t the coordinates given; set f `(x) =-1 and
f ``(x)=-10.

I think d=30 and c=-1.
So I need to solve for a,b.

I didn't think solving for x would be this tricky? What have I missed?

$
f(x) = ax^3+bx^2+cx+d
$

$f(0) = 30$ ... $d = 30$

$
f'(x) = 3ax^2 + 2bx + c
$

$f'(0) = -1$ ... $c = -1$

$f'(1) = -17$ ... $\textcolor{red}{3a + 2b -1 = -17}$

$
f''(x) = 6ax + 2b
$

$f''(1) = -10$ ... $\textcolor{red}{6a+2b = -10}$

solve the two equation system for $a$ and $b$
• March 24th 2010, 08:42 AM
Neverquit
I understand the a, b, c, d (a=2,b=-11,c=-1,d=30) part. It's applying the theory of polynomials equations to solve for the roots. I haven't any experience with this type of problem.
Any information is appreciated.
• March 24th 2010, 10:58 AM
skeeter
Quote:

Originally Posted by Neverquit
I understand the a, b, c, d (a=2,b=-11,c=-1,d=30) part. It's applying the theory of polynomials equations to solve for the roots. I haven't any experience with this type of problem.
Any information is appreciated.

The Rational Roots Test
• March 24th 2010, 08:39 PM
Neverquit
Theory of equations solution?
Hi..
these are the numbers for a=2, b=-11, c=-1, d=30. st these into equation
y=ax^3+bx^2+cx+d gives:

y=2x^3-11bx^2-x+30

If I do the Rational Roots test then I get
+/- 1,2,3,5,10,15,30,1/2,3/2,5/2,15/2

I know the roots are -3/2,5,2 but what does the solution look like with the theory of equations method solving for x1, x2, x3?