Chapter 4.4, Problem 80E

To determine

To solve: the given quadratic equation by factorization.

Answer to Problem 80E

The solution of the given equation is $x=\frac{1}{9},x=3$ .

Explanation of Solution

Given information:

A quadratic equation is given as

  $9x^2-28x+3=0$

Calculation:

Consider the given equation.

  $9x^2-28x+3=0$

Now, break the coefficient of middle term $-28$ as sum or difference of two numbers such that product of the numbers is equal to the product of first term and constant term that is product of $9$ and $3$ .

Such numbers will be $-27$ and $-1$ .

Now, factorize left side of $9x^2-28x+3=0$ as shown:

  $\begin{array}{c}9x^2-28x+3=0\\ 9x^2-27x-x+3=0\\ 9x\left(x-3\right)-1\left(x-3\right)=0\\ \left(9x-1\right)\left(x-3\right)=0\end{array}$

Solution can be found by equating each factor with 0.

  $\begin{array}{c}9x-1=0\\ x=\frac{1}{9}\end{array}$ and $\begin{array}{c}x-3=0\\ x=3\end{array}$

Therefore, the solution of the given equation is $x=\frac{1}{9},x=3$ .