help with solving for x

manutd

solve the following equation for x (correct to 1 decimal place)
5x^3-35x^2+47x=0
do you use the formula ax^3 + bx^2 + cx + d = 0

Bit cynical

Note that in your equation, there's no constant term, i.e. no equivalent to d in the second equation. This suggests that there's an operation you can perform on the left hand of your equation to simplify things.

Is there some common factor in each of the terms on the left hand side that can be factored out?

SomethingElse

x=0

Or

5x^2-35x+47=0

Solving:

x=5.188, 1.812 or x=0

manutd

Bit cynical wrote: »

Note that in your equation, there's no constant term, i.e. no equivalent to d in the second equation.  This suggests that there's an operation you can perform on the left hand of your equation to simplify things.

Is there some common factor in each of the terms on the left hand side that can be factored out?

x is a common factor.

Bit cynical

manutd wrote: »

x is a common factor.

Well you basically have it then. Divide both sides by x and solve the quadratic. x=0 becomes an additional root to the two roots of the quadratic equation.

TheBody

Bit cynical wrote: »

Well you basically have it then. Divide both sides by x and solve the quadratic. x=0 becomes an additional root to the two roots of the quadratic equation.

The correct procedure is to factorise and then solve.

You can't divde both sides by x because you potentially could be dividing by zero.