
Solve $ - 3{x^2} - 5 = 22$ ?
Answer
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Hint: Solve the quadratic equation $ - 3{x^2} - 5 = 22$ by separating the variable and constant term. We get squared of the variable equal to the negative value. When squared gives a negative result then, we introduce an imaginary number.
The solution of the equation is the complex number.
The imaginary number is $i$ , which is the $\sqrt { - 1} $ .
If we solve $ - 3{x^2} - 5 = 22$, we get $ - {x^2} = 9$ .
And, ${x^2} = - 9$
Here, we write, $\sqrt { - 1} = i$ and $\sqrt 9 = \pm 3$ .
Complete step by step answer:
Consider the quadratic equation is given by $ - 3{x^2} - 5 = 22$.
Add $5$ to each side of the equation.
$ - 3{x^2} - 5 + 5 = 22 + 5$
$ \Rightarrow - 3{x^2} = 27$
Divide each side of the equation by $ - 3$ we get,
${x^2} = - 9$
When we square a positive and negative number we get a positive result, but here we get the $x$ squared is $ - 9$ .
Apply square root each side of the equation, ${x^2} = - 9$
$ \Rightarrow x = \sqrt { - 9} $
$ \Rightarrow x = \sqrt { - 1} \times \sqrt 9 $
Here we use the imaginary number $i$ , which is the $\sqrt { - 1} $ .
$ \Rightarrow x = \pm i \times 3$
$ \Rightarrow x = 3i$ and $x = - 3i$
Final Answer: The solution of the equation is $x = 3i$ and $x = - 3i$.
Note: Most of the students make mistakes in solving ${x^2} = - 9$. They simply find the square root of $9$ that is $3$ and put a minus sign before $ - 3$. So the solution is $x = - 3$ which is incorrect.
Remember the following points;
When squared give a negative result so we use complex number,
this doesn't happen, because:
when we square a positive number we get a positive result, and
when we square a negative number we also get a positive result (because a negative times a negative gives a positive), for example \[ - {\mathbf{2}}{\text{ }} \times - {\mathbf{2}} = + {\mathbf{4}}\] .
The solution of the equation is the complex number.
The imaginary number is $i$ , which is the $\sqrt { - 1} $ .
If we solve $ - 3{x^2} - 5 = 22$, we get $ - {x^2} = 9$ .
And, ${x^2} = - 9$
Here, we write, $\sqrt { - 1} = i$ and $\sqrt 9 = \pm 3$ .
Complete step by step answer:
Consider the quadratic equation is given by $ - 3{x^2} - 5 = 22$.
Add $5$ to each side of the equation.
$ - 3{x^2} - 5 + 5 = 22 + 5$
$ \Rightarrow - 3{x^2} = 27$
Divide each side of the equation by $ - 3$ we get,
${x^2} = - 9$
When we square a positive and negative number we get a positive result, but here we get the $x$ squared is $ - 9$ .
Apply square root each side of the equation, ${x^2} = - 9$
$ \Rightarrow x = \sqrt { - 9} $
$ \Rightarrow x = \sqrt { - 1} \times \sqrt 9 $
Here we use the imaginary number $i$ , which is the $\sqrt { - 1} $ .
$ \Rightarrow x = \pm i \times 3$
$ \Rightarrow x = 3i$ and $x = - 3i$
Final Answer: The solution of the equation is $x = 3i$ and $x = - 3i$.
Note: Most of the students make mistakes in solving ${x^2} = - 9$. They simply find the square root of $9$ that is $3$ and put a minus sign before $ - 3$. So the solution is $x = - 3$ which is incorrect.
Remember the following points;
When squared give a negative result so we use complex number,
this doesn't happen, because:
when we square a positive number we get a positive result, and
when we square a negative number we also get a positive result (because a negative times a negative gives a positive), for example \[ - {\mathbf{2}}{\text{ }} \times - {\mathbf{2}} = + {\mathbf{4}}\] .
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