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||\sqrt[\Large{\color{#3b87cd}{\textbf{index}}}]{\color{#ec0000}{\textbf{radicand}}}= \color{#3a9a38}{\textbf{root}}||
Generalizing, we get:
The square root of a number |y| corresponds to a positive real number |x| which, squared, equals |y.| ||\sqrt{y} = x\ \Longleftrightarrow\ x^2=y||
In summary, taking a square root and squaring are inverse operations.
The square root of |49| is |7,| because |7| squared equals |49.| ||\sqrt{49} = 7\ \Longleftrightarrow\ 7^2=49||
A square root can be used to isolate a variable that is squared in an algebraic equation. It's very useful for finding missing measurements in plane figures from their area.
||\begin{align} c^2 &= 184.96\\ \color{#ec0000}{\sqrt{\color{black}{c^2}}}&=\color{#ec0000}{\sqrt{\color{black}{184.96}}}\\ c &= 13.6\end{align}||
In the set of real numbers |(\mathbb{R}),| we can't calculate the square root of negative numbers.
This is because of the sign rule that states that any number squared equals a positive number, regardless if it is positive or negative itself. For example, |5^2= 5 \times 5 = 25| and |(-5)^2=-5 \times -5 = 25| too. Therefore, in |\mathbb{R},| it is impossible to calculate the square root of |-25,| since there is no number that, when multiplied by itself, equals |-25.|
Yes, the set of complex numbers, denoted |(\mathbb{C})| allows us to calculate the square root of a negative number. This set corresponds to the set of all real numbers in addition to an imaginary number, denoted |i,| defined as follows: ||i=\sqrt{-1}\ \Longleftrightarrow\ i^2=-1||With the addition of this one number, which is not a real number, taking the square root of negative numbers becomes possible. Complex numbers, which are not part of the Quebec school curriculum, are widely used in science and engineering to solve equations that would otherwise be impossible to solve.
According to the definition, the square root of a number is always positive. However, when solving second degree equations, it's important to consider both solutions, not just the positive one.
In other words, even if the square root of |81| is |9| (positive), the equation |x^2=81| has 2 solutions: |9| and |-9.|
What are the solutions of the following equation? ||8x^2=450||
First divide by |8| to isolate |x^2.| ||\begin{align} 8x^2&=450\\ \color{#ec0000}{\dfrac{\color{black}{8x^2}}{\boldsymbol{8}}} &= \color{#ec0000}{\dfrac{\color{black}{450}}{\boldsymbol{8}}}\\ x^2&=56.25\end{align}||We are now looking for a number |x| which, when squared, equals |56.25.| We must therefore calculate the square root of |56.25.| ||\sqrt{56.25}=7.5|||7.5| is a solution of the equation. However, it's not the only solution, since |-7.5| is also a valid solution. The following validates both of these solutions.||\begin{gather}8\boldsymbol{\color{#3a9a38}{x}}^2=450\\\swarrow\searrow\\\begin{aligned}8(\boldsymbol{\color{#3a9a38}{7.5}})^2&\overset{?}{=}450 &8(\boldsymbol{\color{#3a9a38}{-7.5}})^2&\overset{?}{=}450\\[3pt]8(56.25)&\overset{?}{=}450&8(56.25)&\overset{?}{=}450\\[3pt]450&=450&450&=450\end{aligned}\end{gather}||Answer: The solutions of the equation are |7.5| and |-7.5.|
Whenever we solve a quadratic equation, to avoid forgetting the 2nd solution, we use the symbol |\pm| when we take the square root.||\begin{aligned} x^2&=56.25\\ x&=\color{#ec0000}{\boldsymbol\pm\sqrt{\color{black}{56.25}}}\\ &\swarrow\ \searrow\end{aligned}\\ \begin{alignat}{1}\!\!\!\!\!\! x_1 &=\color{#ec0000}{\boldsymbol +\sqrt{\color{black}{56.25}}} \qquad x_2&=\color{#ec0000}{\boldsymbol -\sqrt{\color{black}{56.25}}}\\ &=\color{#ec0000}{\boldsymbol +}\,7.5 &=\color{#ec0000}{\boldsymbol -}\,7.5 \end{alignat}||