Question

Show that the equation

2x^3-3x^2+7x-18=0 has a real root which is real and positive

Answer 1

Solution;

$2x^3-3x^2+7x-18=0$

The standard form of a quadratic equation is ;

$Ax^3+Bx^2+Cx+D=0$

Base on our equation;

A=2

B=-3

C=7

D=-18

The first step,check if x=1 is a solution of the equation;

If A+B+C+D=0,then x=1 is a solution.

From the equation;

2-3+7-18=-12

Hence x=1 is not a solution of the equation.

The next step check if x=-1 is a solution,

If the sum of alternative coefficient is the same;

A+C=2+7=9

B+D=-3-18=-21

Hence;

$9\neq-21$ so x=-1 is not a solution of the problem.

The next step is try x=2 using synthetic division;

$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} 2 & 2&-3 &7&-18 \\ \hline & 2 & 1&9&0 \\ \hdashline & x^2 & x \end{array}$

The remainder R=0

Hence x=2 is a solution of the problem.