Question

How do solve \[2{{x}^{2}}+18=0\] ?

Answer 1

Hint: We will solve the above question by using the formula \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\] . First, we will put the values in the above equation from the given question according to the general quadratic equation. Then, we will find the values of x.

Complete step by step answer:
Let us solve the question.
We have given the equation as \[2{{x}^{2}}+18=0\] .
As we know that the general form of the quadratic equation is \[a{{x}^{2}}+bx+c=0\] .
And the values of x according to the equation \[a{{x}^{2}}+bx+c=0\] are \[\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\] .
According to the above equations, we can say that
a=2, b=0, and c=18.
First, we will check if the roots are real or not.
For that, we have to find the discriminant or D.
\[D=\sqrt{{{b}^{2}}-4ac}=\sqrt{{{0}^{2}}-4\times 2\times 18}=12\times \sqrt{-1}\]
As D is not real, then the roots of the equation \[2{{x}^{2}}+18=0\] are also not real.
Therefore, the roots of the equation are imaginary.
Now, we will put the values of a, b, and c in the equation \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\] , we get
\[x=\dfrac{-0\pm \sqrt{{{0}^{2}}-4\times 2\times (18)}}{2\times 2}\]
Now, we will simplify the above equation, we get
\[x=\dfrac{\pm \sqrt{-144}}{2\times 2}=\pm \dfrac{12\sqrt{-1}}{4}=\pm 3\sqrt{-1}\]
Let us say \[\sqrt{-1}=i=iota\] .

Then, the values of x are \[3i\] and \[-3i\]

Note: In this type of question, we should have a proper knowledge in quadratic equation. We can solve this by a second method simply. Suppose, we have the equation \[2{{x}^{2}}+18=0\] .
We will take the number 18 to the right side of the equation. After that, we will find the value of x.
\[\Rightarrow {{x}^{2}}=-\dfrac{18}{2}\]
Now, by taking the square root both sides of the above equation, we get
\[\Rightarrow x=\sqrt{-\dfrac{18}{2}}\]
\[\Rightarrow x=\sqrt{-9}\]
\[\Rightarrow x=\pm 3\sqrt{-1}\]
As there is -1 inside the root, then we will put the value of \[\sqrt{-1}\] as \[i\] .
Therefore, we get the values of x as \[3i\] and \[-3i\] .