Chapter 1.3, Problem 27AYU

In Problems 17-36, solve each equation algebraically. Verify your solution using a graphing utility.

$x^2-3x-28=0$

To determine

To find: The solution for the equation $x^2-3x-28=0$ algebraically and verify using a graphing utility.

Answer to Problem 27AYU

Solution:

$x\text{ = −4, 7}$

Explanation of Solution

Given:

The equation $x^2-3x-28=0$

Algebraic solution:

$x^2-3x-28=0$

$x^2-7x+4x-28=0$Splitting the middle term

$x\left(x\text{ − 7}\right)\text{ + 4}\left(x\text{ − 7}\right)\text{ = 0}$Taking common factor $x$ from first two terms and 4 from next two term

$\left(x\text{ − 7}\right)\left(x\text{ + 4}\right)\text{ = 0}$

$x\text{ − 7 = 0}$ or $x\text{ + 4 = 0 }$

$x\text{ = 7}$ or $x\text{ = −4}$

Check:

Let $x\text{ = 7}$ in the expression in $x$ on the left side of the equation

$\left(x\text{ + 2}\right)\left(x\text{ − 3}\right)\text{ = }(\text{3 + 2}) \left(\text{3 − 3}\right)$

$\text{= 5 × 0}$

$\text{= 0}$

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

${(x - 3)}^2={(3 - 3)}^2$

$\text{= 0}$

$\text{= 0}$

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