Question

How do you solve \[2{{x}^{2}}+12=0\]?

Answer 1

Hint: This type of problem is based on the concept of quadratic equation. Here, we first consider the equation \[2{{x}^{2}}+12=0\]. Then, we have to subtract the whole equation by 12 and cancel 12 from the LHS. And then divide the obtained equation by 2. Make some necessary calculations and find the value of \[{{x}^{2}}\]. Take the square root on both the sides of the equation and find the value of x which is the required answer.

Complete step by step solution:
According to the question, we are asked to solve the equation \[2{{x}^{2}}+12=0\].
We have been given the equation is \[2{{x}^{2}}+12=0\]. --------(1)
We first have to consider the equation (1).
Subtract the whole equation by 12.
\[\Rightarrow 2{{x}^{2}}+12-12=0-12\]
We know that terms with the same magnitude and opposite signs cancel out. On cancelling 12, we get
\[2{{x}^{2}}=0-12\]
On further simplifications, we get
\[2{{x}^{2}}=-12\]
Now, we have to divide the whole equation by 2.
\[\Rightarrow \dfrac{2{{x}^{2}}}{2}=\dfrac{-12}{2}\]
We can express the equation as
\[\dfrac{2{{x}^{2}}}{2}=\dfrac{-2\times 6}{2}\]
We find that 2 are common in both the numerator and denominator of LHS and RHS. On cancelling 2, we get
\[{{x}^{2}}=-6\]
We need to find the value of x.
So we have to take square root on both sides.
\[\Rightarrow \sqrt{{{x}^{2}}}=\sqrt{-6}\]
But we know that \[\sqrt{{{x}^{2}}}=\pm x\].
Therefore, we get
\[x=\pm \sqrt{-6}\]
We know that \[\sqrt{ab}=\sqrt{a}\cdot \sqrt{b}\]. Using this property of roots, we get
\[x=\pm \sqrt{-1}.\sqrt{6}\]
We know that \[\sqrt{-1}=i\], where i is a complex number.
We get
\[x=\pm i\sqrt{6}\]

Therefore, the values of x in the given equation \[2{{x}^{2}}+12=0\] are \[i\sqrt{6}\] and \[-i\sqrt{6}\].

Note: We should know that a negative term under a square root is always imaginary value. Avoid calculation mistakes based on sign conventions. Since the given equation is quadratic, we will get two values of x. We can also solve this question using quadratic formulas.