Question

In the complex number system, solve the equation

$9x^4+33x^3-71x^2-57x-10=0$

Short answer

The solution to the given equation are$x=2,-5,-\frac{1}{3},-\frac{1}{3}$

Step-by-step solution

Step 1. Given Information

The given equation is$9x^4+33x^3-71x^2-57x-10=0$

Step 2. Explanation

The given equation is a fourth power equation in x which can be solved by arranging the given equation in such a way that the equation forms a product of two quadratic equation.

The coefficient of $x^4$is 9 which is a perfect square of 3.

This is possible when the given equation is the product of quadratic equations of the form $3x^2+ax+band3x^2+cx+d$

Let us compute the product of quadratic equations.

$(3x^2+ax+b)(3x^2+cx+d)=9x^4+3(a+c)x^3+(3b+3d+ac)x^2+(ad+bc)x+bd$

On comparing the above equation with the given equation, we get,

role="math" localid="1648195521568" $$\begin{gathered} a+c=11--\left(1\right) \\ 3b+3d+ac=-71--\left(2\right) \\ ad+bc=-57--\left(3\right) \\ bd=-10--\left(4\right) \end{gathered}$$

Let us consider the last relationship which is possible only if either role="math" localid="1648195068578" $b=2andd=-5orb=-2andd=5$

Next consider the second equation obtained above, on putting the values of b as 2 and d as$-5$, we get the product of a and c as$-62$which is not satisfied by the first equation which says the sum of a and c is 11

Step 3. Calculation

Substitute the values of $b=-2,d=5$in third equation,

$$\begin{gathered} ad+bc=-57 \\ a\left(5\right)+(-2)c=-57 \\ 5a-2c=-57 \end{gathered}$$

Solve the above obtained equation and the first equation simultaneously,

$$\begin{gathered} a=11-c \\ 5a-2c=-57 \\ 5(11-c)-2c=-57 \\ 7c=112 \\ c=16 \end{gathered}$$

Find the value of a by substituting the value of c,

$$\begin{gathered} a=11-c \\ a=11-16 \\ a=-5 \end{gathered}$$

Step 4. Calculation

Thus, the quadratic equations for which the product is the given equation are given below,

$$\begin{gathered} 3x^2-5x-2 \\ 3x^2+16x+5 \end{gathered}$$

Solve the above two quadratic equations for x.

$$\begin{gathered} 3x^2-5x-2=0 \\ x=\frac{5\pm \sqrt{{(-5)}^2-4\left(3\right)(-2)}}{2\left(3\right)} \\ x=\frac{5\pm 7}{6} \\ x=2,-\frac{1}{3} \\ And \\ 3x^2+16x+5=0 \\ x=\frac{-16\pm \sqrt{{\left(16\right)}^2-4\left(3\right)\left(5\right)}}{2\left(3\right)} \\ x=\frac{-16\pm 14}{6} \\ x=-5,-\frac{1}{3} \end{gathered}$$