Chapter 1.3, Problem 30AYU

To determine

To find: The solution for the equation $5x^2 = 13x+6$ algebraically and verify using a graphing utility.

Answer to Problem 30AYU

Solution:

$x\text{ = 3, }\frac{-2}{5}$

Explanation of Solution

Given:

The equation $5x^2\text{ = 13}x+6$

Algebraic solution:

$5x^2 = 13x+6$

$5x^2 -13x=13x+6-13x$Subtract $\text{13}x$ from both sides

$5x^2\text{ − 13}x\text{ − 6 = 6 − 6}$Subtract 6 from both sides

$\text{5}{x}^2\text{ − 13}x\text{ − 6 = 0 }$

$5x^2\text{ − 15}x\text{ + 2}x\text{ − 6 = 0}$Splitting the middle term

$\text{5}x\left(x\text{ − 3}\right)\text{ + 2}\left(x\text{ − 3}\right)\text{ = 0}$Taking $\text{5}x$ common from first two terms and 2 common from last two terms

$\left(x\text{ − 3}\right)\left(\text{5}x\text{ + 2}\right)\text{ = 0}$Taking $(x\text{ − }3)$ common factor from the terms

$x=3$      $5x=-2$

$x = \frac{-2}{5}$

Check:

Let $x\text{ = 3}$ in the expression in $x$ on the left side of the equation

$5{(3)}^2 =5\times 9=45$

Let $x\text{ = 3}$ in the expression in $x$ on the right side of the equation

$\text{13}\left(\text{3}\right)\text{ + 6 = 39 + 6}$

$\text{= 45}$

Since the expressions are equal, the solution $x\text{ = 3}$ checks

Let $x = \frac{-2}{5}$ in the expression in $x$ on the left side of the equation

$5{(\frac{-2}{5})}^2 = 5\times {(\frac{-2}{5})}^2$

$= \frac{4}{5}$

Let $x = \frac{-2}{5}$ in the expression in $x$ on the right side of the equation

$13(\frac{-2}{5}) + 6 = \frac{-26}{5} + 6$

$= \frac{-26+30}{5}$   Take LCM

$= \frac{4}{5}$

Since the two expressions are equal, the solution $x = \frac{-2}{5}$ checks.