Question

How do you solve \[3{{y}^{2}}+2y-1=0~\]

Answer 1

Hint: We are given \[3{{y}^{2}}+2y-1=0~\] to solve this we learn about the type of Equation we are given then learn number of solutions of equation. We will learn how to factor the quadratic equation; we will use the middle term split to factor the term and we will simplify by taking common terms out. We also use zero product rules to get our answer. To be sure about your answer we can also check by putting the acquired value of the solution in the given Equation and check whether they are the same or not .

Complete step-by-step solution:
We are given, we are asked to solve the given problem \[3{{y}^{2}}+2y-1=0~\]. First, we observe that it has a maximum power of ‘$2$’ so it is a quadratic equation. Now we should know that a quadratic equation has a $2$ solution or we say an equation of power ‘n’ will have an ‘n’ solution.
Now as it is a quadratic equation, we will change it into standard form \[a{{y}^{2}}+by+c=0\]. As we look closely our problem is already in standard form
\[3{{y}^{2}}+2y-1=0~\]
Now we have to solve the equation \[3{{y}^{2}}+2y-1=0~\]
To solve this equation, we first take the greatest common factor possibly available to the terms. As we can see that in \[3{{y}^{2}}+2y-1=0~\]
$3,2,-1$ has nothing in common.
So, Equation remains same
\[3{{y}^{2}}+2y-1=0~\]
Now we will solve this using method which is called by factoring, We factor using middle term split For quadratic equations \[a{{y}^{2}}+by+c=0\] we will find such a pair of term whose products is same. As $a\times c$ and whose sum or difference will be equal to the b
Now in \[3{{y}^{2}}+2y-1=0~\]
We have \[a=3\text{ }b=2\text{ and }c=-1~\]
So $a\times c=3$\times$ -1 = -3~$
We can see that \[3\times -1= -3\] and
Their sum is \[3-1=2~\]
So, we use this to split
 \[3{{y}^{2}}+2y-1=0~\]
\[3{{y}^{2}}+\left( 3-1 \right)y-1=0~\]
Opening brackets, we get
\[3{{y}^{2}}+3y-y-1~\]
Taking common in first two and last two terms
We get
\[3y\left( y+1 \right)-1\left( y+1 \right)=0~\]
As y+1 is common so Simplifying further
\[\left( 3y-1 \right)\left( y+1 \right)~\]
Using zero product rule which says if two terms product is zero that either one of them is zero So either \[3y-1=0\text{ }or\text{ }y+1=0~\]
So, we get \[3y=1\text{ }and\text{ }y =-1~\]
Hence solution is \[y=\dfrac{1}{3}\text{ and }y=-1~\]

Note: We have a quadratic equation \[3{{y}^{2}}+2y-1=0~\]we can easily solve it using the quadratic formula given as for an equation: $a{{y}^{2}}+by+c=0 $. Solution is given as : ${{y}^{2}}=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
So in our quadratic equation \[3{{y}^{2}}+2y-1=0 \]
We have, \[a=3\text{ }b=2\text{ and }c=-1 \]
Using this in above formula we get
\[\]${{y}^{2}}=\dfrac{-2\pm \sqrt{{{2}^{2}}-4\times 3\times (-1)}}{2(3)}$
Simplifying we get
${{y}^{2}}=\dfrac{-2\pm \sqrt{16}}{6}$
As \[{{2}^{2}}-4\times 3\times (-1)=16\]
So, we get
\[y=\dfrac{-2+4}{6}\,\,\,\,\,\text{and}\,\,\,y=\dfrac{-2-4}{6}\]
Simplifying we get
\[y=\dfrac{2}{3}=\dfrac{1}{3}\,\,\,\,\,\text{and}\,\,x=\dfrac{-6}{6}=-1\]
So, $y=\dfrac{1}{3}\,\,\,\,\,and\,\,\,y=-1$ is the solution.