Question

Solve the following equation: ${{t}^{2}}-49=0$ .

Answer 1

Hint: As 49 is the square of 7, so start by writing the equation as ${{t}^{2}}-{{7}^{2}}=0$ followed by using the formula ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ to convert it to $\left( t+7 \right)\left( t-7 \right)=0$ . For the equation $\left( t+7 \right)\left( t-7 \right)=0$ to hold true, at least one of the two terms, i.e., (t+7) and (t-7) must be zero. So, you need to solve the equation t+7=0 and t-7=0 separately and take the union of values you get in each case, to get the final answer.

Complete step-by-step answer:

The given equation is ${{t}^{2}}-49=0$ and we know that 49 is the square of 7. So, our equation becomes:
${{t}^{2}}-{{7}^{2}}=0$
We also know that ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ . So, our equation turns out to be
$\left( t+7 \right)\left( t-7 \right)=0$
The equation given is $\left( t+7 \right)\left( t-7 \right)=0$ , and for these equations to be satisfied there are two possible cases. Either (t+7) is equal to zero, else (t-7) is equal to zero.
So, let us find the values of t for which (t+2) is equal to zero.
$t+7=0$
$t=-7$
Also, for the value t=-7, t-7=-14, which is finite, so the equation $\left( t+7 \right)\left( t-7 \right)=0$ is satisfied for t=-7.
Now, we will try to find the values of t for which t-7=0.
$t-7=0$
$\Rightarrow t=7$
Also, for the value t=7, t+7=14, which is finite, so the equation $\left( t+7 \right)\left( t-7 \right)=0$ is satisfied for t=7 as well.
So, all the possible values of t are 7 and -7. , i.e., $x\in \{7,-7\}$ .

Note: You need to remember the formula ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ for solving such questions. You could have also solved the equation by directly taking 49 to the other side of the equation and solving the equation.
${{t}^{2}}-49=0$
$\begin{align}
  & \Rightarrow {{t}^{2}}=49 \\
 & \Rightarrow t=\pm \sqrt{49}=\pm 7 \\
\end{align}$