Question

Solve the following equation for y, $9{{y}^{2}}+5=0$.

Answer 1

Hint: To solve this question, we should know that $\sqrt{-1}=i$ and we should also know that $\sqrt{{{a}^{2}}}=\pm a$. We will apply multiple arithmetic operations to keep the variables on one side and all the rest values on the other side and then we will use the above-mentioned properties to get the answers.

Complete step-by-step solution -
In this question, we have been asked to find the value of y for the given equation, $9{{y}^{2}}+5=0$. To solve this question, we will first try to keep the variables on one side and all the rest values on the other side. So, to get that, we will subtract 5 from both sides of the equation. So, we get,
 $9{{y}^{2}}+5-5=0-5$
And we can further write the equation as,
$9{{y}^{2}}=-5$
Now, we will divide both sides of the equation by 9. So, by doing so, we get,
$\dfrac{9{{y}^{2}}}{9}=\dfrac{-5}{9}$
Now, we know that the common terms of the numerator and the denominator gets canceled out. So, we can write the equation as,
${{y}^{2}}=\dfrac{-5}{9}$
Now, we will take the square root of both sides of the equation. So, we get,
$\sqrt{{{y}^{2}}}=\sqrt{\dfrac{-5}{9}}$
Which can be further written as, $y=\pm \sqrt{\dfrac{-5}{9}}$, because $\sqrt{{{a}^{2}}}=\pm a$.
Now, we will further simplify it as,
$y=\pm \sqrt{-1}\times \sqrt{\dfrac{5}{9}}$
We know that $\sqrt{-1}=i$, so we can write the equality as,
$\begin{align}
  & y=\pm \sqrt{\dfrac{5}{9}}i \\
 & \Rightarrow y=\pm \dfrac{\sqrt{5}}{\sqrt{9}}i \\
 & \Rightarrow y=\pm \dfrac{\sqrt{5}}{3}i \\
\end{align}$
Hence, we can say that y has imaginary values, which is, $y=\pm \dfrac{\sqrt{5}}{3}i$.

Note: We can also solve this question by using Shridharacharya’s formula, that is, for a quadratic equation $a{{x}^{2}}+bx+c=0$; $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. For our equation, we will put x = y, a = 9, b = 0, and c = 5 and then we will simplify. So, we will get the answer as, $y=\pm \dfrac{\sqrt{5}}{3}i$.