Math Is Fun Forum

PhuongMath

Member

Registered: 2014-10-25

Posts: 3

Solve Equation

a. $7x^3+6x^2+12x+8=0$
b. $x^3-3x^2+3(\sqrt{2}+1)x-3-2\sqrt{2}=0$

Bob

Administrator

Registered: 2010-06-20

Posts: 10,134

Re: Solve Equation

hi PhuongMath

Welcome to the forum. 

I couldn't spot an easy factorisation, so I went to the function grapher: http://www.mathsisfun.com/data/function … x)&func2=2

Both cubics have only one intersection with the x axis and zooming in shows it's not an integer.

So what methods do you know for getting a value for this ? eg. Newton's (approximation) method

Once you have one factor you could go on to find two complex values but it will not be easy arithmetic.

Bob

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Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

PhuongMath

Member

Registered: 2014-10-25

Posts: 3

Re: Solve Equation

1.
(1) $\Leftrightarrow (x+2)^3=-6x^3\Leftrightarrow x+2=x\sqrt[3]{-6}\Leftrightarrow x=\frac{2}{\sqrt[3]{-6}-1}$
I have not solved the 2nd equation.
You could help me.

Bob

Administrator

Registered: 2010-06-20

Posts: 10,134

Re: Solve Equation

hi PhuongMath

gives

Yes, that is consistent with my graphing attempt.  x is approximately -0.70994

Cube root (6) = 1.817121  = r

The three cube roots of -6 are therefore -1.817121, and complex solutions with the same modulus, r and arguments, + or - pi/3

This places them 1/3 of a turn apart on a circle radius r.

So you can construct the other two values of x from this.

If this is not what you were expecting, then you'll need to go back to the start of the problem so I can see how you got to that equation.

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob