Chapter 10, Problem 1MCT

Solve.

$x^2=49$

To determine

Solve $x^2=49$.

Answer to Problem 1MCT

The two solutions are: $x=7\&x=-7$.

Explanation of Solution

Given:

The given equation is: $x^2=49$.

Concept Used:

The equation will be converted into the form: $x^2-a^2=0$.

Then, we will factorize $x^2-a^2$ into $(x+a)(x-a)$ as,

  $\begin{array}{l}{x}^2-a^2\\ =x^2-ax+ax-a^2\\ =x^2-xa+ax-a^2\\ =x(x-a)+a(x-a)\\ =(x+a)(x-a)\end{array}$

Now, we get the equation: $(x+a)(x-a)=0$

Here, a product of several terms equals to zero.

When a product of two more terms equals to zero, then at least one term will be equal to zero.

So, we can equal the terms to zero: $x-a=0,x+a=0$

These are two equations of single variable.

Simply solving them by addition and subtraction:

  $\begin{array}{l}x-a=0\\ \Rightarrow x-a+a=0+a\\ \Rightarrow x=a\end{array}$

And,

  $\begin{array}{l}x+a=0\\ \Rightarrow x+a-a=0-a\\ \Rightarrow x=-a\end{array}$

Calculation/Explanation:

  $\begin{array}{l}{x}^2=49\\ \Rightarrow x^2-49=0\\ \Rightarrow x^2-7^2=0\\ \Rightarrow (x+7)(x-7)=0\\ \Rightarrow x+7=0;x-7=0\\ \Rightarrow x=-7;x=7\end{array}$

Conclusion:

The solutions of the equation is: $x=7,x=-7$.