Question

How do you find the real solutions of the polynomial $9{x^3} = 5{x^2} + x$?

Answer 1

Hint: According to the question we have to find the real solutions of the polynomial $9{x^3} = 5{x^2} + x$ which is as mentioned in the question. So, first of all to determine the required real solutions of the polynomial we have to take x as a common term in the expression.
Now, on solving the expression we can easily obtain one root or solution of the expression.
Now, we have to obtain the values of a, b and c which are the coefficient of ${x^2}$, coefficient of x and the constant term respectively.
Now, we have to solve the quadratic expression which is as obtained after dividing the expression by x with the help of the quadratic expression to determine the roots of the expression which is as mentioned below:

Formula used: $ x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}......................(A)$
Where, a, is the coefficient of ${x^2}$, b is the coefficient of x and c is the constant term.
Now, we have to solve the expression by adding and subtracting the terms of the expression to determine the roots/zeroes or we can say that the solution of the quadratic expression.

Complete step-by-step solution:
Step 1: First of all to determine the required real solutions of the polynomial we have to take x as a common term in the expression.
$ \Rightarrow x(9{x^2} - 5x - 1) = 0$
Step 2: Now, on solving the expression we can easily obtain one root or solution of the expression. Hence,
$ \Rightarrow x = 0$ and,
$ \Rightarrow (9{x^2} - 5x - 1) = 0$
Step 3: Now, we have to obtain the values of a, b and c which are the coefficient of${x^2}$, coefficient of x and the constant term respectively. Hence,
$ \Rightarrow a = 9,b = - 5$ and
$ \Rightarrow c = 1$
Step 4: Now, we have to solve the quadratic expression which is as obtained after taking x as a common term by the help of the quadratic expression to determine the roots of the expression which is as mentioned in the solution hint. Hence,
$ \Rightarrow x = \dfrac{{ - ( - 5) \pm \sqrt {{{( - 5)}^2} - 4 \times 9 \times ( - 1)} }}{{2 \times 9}}$
Step 5: Now, we have to solve the expression by adding and subtracting the terms of the expression to determine the roots/zeroes or we can say that the solution of the quadratic expression. Hence,
\[
   \Rightarrow x = \dfrac{{5 \pm \sqrt {25 + 36} }}{{18}} \\
   \Rightarrow x = \dfrac{{5 \pm \sqrt {61} }}{{18}}
 \]

Hence, with the help of the formula (A) we have determined the required real solutions of the expression which are \[x = \dfrac{{5 \pm \sqrt {61} }}{{18}}\] and $x = 0.$

Note: To obtain the one of the real solutions of the given expression it is necessary that we have to take x as a common term and then we have to solve it with the right hand side number which is 0.
To determine the two real solutions of the quadratic expression it is necessary that we have to compare the quadratic expression with the general form of the quadratic expression which is $a{x^2} + bx + c = 0$ to obtain the values of a, b, and c which are the coefficient of ${x^2}$, x and the constant term respectively.