Question

How do you solve $\left( {x - 7} \right)\left( {x + 2} \right) = 0$ using the zero-product property?

Answer 1

Hint: In this question, we are going to solve the given quadratic function by using the zero product property.
Given that the product of two factors is equal to zero.
Now we are going to compare the given product of two factors with the zero product property, while comparing these we can write the product of the first factor equal to zero and the product of the second factor equal to zero.
While equating those terms we can get the required values
Finally, we can get the required solution by using the zero product property.

Formula used: The zero product property states that if $a \cdot b = 0$ then either $a$ or $b$ equals zero.

Complete step-by-step solution:
First, consider the given quadratic function expressed as the product of two factors
$\left( {x - a} \right)\left( {x - b} \right)$ with $\left\{ {a,b} \right\} \in Q$
Mark the given equation as $\left( 1 \right)$
$\left( {x - 7} \right)\left( {x + 2} \right) = 0$ …. $\left( 1 \right)$
The above product is equal to zero. That is either $\left( {x - a} \right) = 0$ or $\left( {x - b} \right) = 0$ …. $\left( 2 \right)$
By comparing the above two equations $\left( 1 \right)$ and $\left( 2 \right)$ we can easily get the values of $a$ and $b$
Here we get the values of $a$ and $b$ as follows
$a = 7$ and $b = - 2$
Hence the zeros of $\left( {x - 7} \right)\left( {x + 2} \right)$ are either $7$ or $ - 2$

Hence, we can get the result as either $x = 7$ or $x = - 2$.

Note: When given any quadratic function for which you wish to find the zeros it is a good start to see if it factorizes in this way.
Now let us solve this question by another method as follows
Let’s solve the equation by using a quadratic formula.
 $\left( {x - 7} \right)\left( {x + 2} \right) = 0$
Simplify both sides of the equation.
${x^2} + 2x - 7x - 14 = 0$
${x^2} - 5x - 14 = 0$
From this equation we can get $a = 1,\,b = - 5,\,c = - 14$
Now use the quadratic formula with $a = 1,\,b = - 5,\,c = - 14$
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Putting the values,
$\Rightarrow$$x = \dfrac{{ - \left( { - 5} \right) \pm \sqrt {{{\left( { - 5} \right)}^2} - 4\left( 1 \right)\left( { - 14} \right)} }}{{2\left( 1 \right)}}$
Simplifying, we get,
$\Rightarrow$$x = \dfrac{{\left( 5 \right) \pm \sqrt {\left( {25} \right) + 56} }}{2}$
Simplifying further,
$\Rightarrow$$x = \dfrac{{\left( 5 \right) \pm \sqrt {81} }}{2}$
$\Rightarrow$$x = \dfrac{{\left( 5 \right) \pm 9}}{2}$
$\Rightarrow$$x = \dfrac{{5 + 9}}{2}$
Finally, we get the required values,
$\Rightarrow$$x = \dfrac{{14}}{2}$
$\Rightarrow$$x = 7$
Or $x = \dfrac{{5 - 9}}{2}$
$\Rightarrow$$x = \dfrac{{ - 4}}{2}$
$\Rightarrow$$x = - 2$
$\Rightarrow$$x = 7$ or $x = - 2$ .