Question

How do you solve \[-3{{n}^{2}}+5n-2=0\] using the formula?

Answer 1

Hint: In this problem, we have to solve and find the value of n. Since the given equation is a quadratic equation, we can use the quadratic formula to find the value of n. we can first compare the given equation and the general quadratic equation to find the values of a, b, c and we can substitute those values in the quadratic formula to get the value of n.

Complete step by step answer:
We know that the given equation to be solved is,
\[-3{{n}^{2}}+5n-2=0\]
We also know that a quadratic equation in standard form is,
\[a{{n}^{2}}+bn+c=0\] ……. (2)
We can now compare the two equations (1) and (2), we get
a = -3, b = 5, c = -2.
We know that the quadratic formula for the standard form \[a{{n}^{2}}+bn+c=0\] is
\[n=\dfrac{-b\pm \sqrt{{{\left( b \right)}^{2}}-4\times a\times \left( c \right)}}{2a}\]
Now we can substitute the value of a, b, c in the above formula, we get
\[\Rightarrow n=\dfrac{-\left( 5 \right)\pm \sqrt{{{\left( 5 \right)}^{2}}-4\times -3\times \left( -2 \right)}}{2\times -3}\]
Now we can simplify the above step, we get
\[\begin{align}
  & \Rightarrow n=\dfrac{\left( -4 \right)\pm \sqrt{25-24}}{-6} \\
 & \Rightarrow n=\dfrac{-5\pm \sqrt{1}}{-6} \\
 & \Rightarrow n=\dfrac{-5\pm 1}{-6} \\
\end{align}\]
Now we can separate the terms to simplify it,
\[\begin{align}
  & \Rightarrow n=\dfrac{-5+1}{-6}=1 \\
 & \Rightarrow n=\dfrac{-5+1}{-6}=\dfrac{2}{3} \\
\end{align}\]
Therefore, the value of \[n=1\], \[n=\dfrac{2}{3}\] .

Note:
Students should always concentrate in the quadratic formula part, where we have to remember the formula and we have to find the values to be substituted in it. We also make mistakes while simplifying the terms inside and outside the root and separating it as it has plus or minus signs in it. We can also substitute the resulted value in the equation to check for the correct values,
We can substitute \[n=1\], \[n=\dfrac{2}{3}\] in equation (1), we get
When n = 1,
\[\begin{align}
  & \Rightarrow -3+5-2=0 \\
 & \Rightarrow -5+5=0 \\
\end{align}\]
When \[n=\dfrac{2}{3}\] ,
\[\begin{align}
  & \Rightarrow -3\times {{\left( \dfrac{2}{3} \right)}^{2}}+5\left( \dfrac{2}{3} \right)-2=0 \\
 & \Rightarrow -\dfrac{4}{3}+\dfrac{4}{3}=0 \\
 & \Rightarrow \\
\end{align}\]
Therefore, we can verify that the value for \[n=1\], \[n=\dfrac{2}{3}\] .