Question

Solve the following equation for x $\left( 13+x \right)\left( 13-x \right)=0$ .

Answer 1

Hint: For the equation $\left( 13+x \right)\left( 13-x \right)=0$ to hold true, at least one of the two terms, i.e., (13+x) and (13-x) must be zero. So, you need to solve the equation 13+x=0 and 13-x=0 separately and take the union of values you get in each case, to get the final answer.

Complete step-by-step answer:

The equation given is $\left( 13+x \right)\left( 13-x \right)=0$ , and for these equations to be satisfied there are two possible cases. Either (13+x) is equal to zero, else (13-x) is equal to zero.
So, let us find the values of x for which (13+x) is equal to zero.
$13+x=0$
$x=-13$
Also, for the value x=-13, 13-x=26, which is finite, so the equation $\left( 13+x \right)\left( 13-x \right)=0$ is satisfied for x=-13.
Now, we will try to find the values of x for which 13-x=0.
$13-x=0$
$\Rightarrow 13=x$
Also, for the value x=13, 13+x=26, which is finite, so the equation $\left( 13+x \right)\left( 13-x \right)=0$ is satisfied for x=13 as well.
So, all the possible values of x are 13 and -13. , i.e., $x\in \{-13,13\}$ .

Note: Be careful as zero multiplied by any number is not 0, but actually zero multiplied by any finite real number is zero. If we multiply zero with infinity, we can get a finite, infinite or zero as the answer. So, it is necessary for the product of two real numbers to be zero that if one is zero the other must be finite.