Chapter 10.5, Problem 2MRE

To determine

To Solve:

  $x^3-7x+6=0$

Answer to Problem 2MRE

The roots of the equation are 1, 2, -3.

Explanation of Solution

Given:

The equation is :

  $x^3-7x+6=0$

Formula Used:

The cubic equation $a{x}^3+b{x}^2+cx+d=0$ has roots p, q ,r ; then ,

  $p+q+r=-\frac{b}{a}$

  $pq+qr+pr=\frac{c}{a}$

  $pqr=-\frac{d}{a}$

Calculation:

Comparing the given equation with the standard cubic equation $a{x}^3+b{x}^2+cx+d=0$ ,we have:

a = 1, b = 0, c = -7 , d = 6

Since the equation is cubic, it has 3 roots.

Let the roots are p, q, r.

Use the $\frac{d}{a}$ method to find the roots of the equation.

Since $\text{product of the roots = pqr =}\frac{-d}{a}=-\mathrm{6...........}(1)$ , the roots of the equation can be $\pm 1,\pm 2,\pm 3$ .

  $p+q+r=-\frac{b}{a}=\mathrm{0.............}(2)$

As x = 1 satisfies the equation:

  $\begin{array}{l}{x}^3-7x+6=0\\ $ {(1)}^3-7(1)+6=0$\end{array}$

So, p = 1 is a root of the equation.

Substitute the value of p = 1 in the equations (1) and (2):

  $\begin{array}{l}(1)qr=-6\Rightarrow qr=-6\\ 1+q+r=0\Rightarrow q+r=-1\end{array}$

Solve the equations:

  $\begin{array}{l}q+r=-1\Rightarrow q=-1-r\\ and\\ qr=-6\Rightarrow (-1-r)r=-6\\ \Rightarrow (1+r)r=\mathrm{6...................}[\text{multiply by -1 each side]}\\ \Rightarrow r^2+r-6=\mathrm{0................}[\text{subtract 6 each side]}\\ \Rightarrow r^2+3r-2r-6=\mathrm{0.......}[\text{factor the middle term]}\\ \Rightarrow r(r+3)-2(r+3)=0\\ \Rightarrow (r+3)(r-2)=0\\ \Rightarrow r=-3\text{ or }r=\mathrm{2..........}[ab=0\iff a=0\text{ or }b=0]\\ \Rightarrow q=-1-(-3)\text{ or }q=-1-\mathrm{2...}[q=-1-r]\\ \Rightarrow q=2\text{ or }q=-3\end{array}$

So, the roots of the equation are 1, 2, -3.