Question

Solve the following for the value of y: $7y=-3{{y}^{2}}-4$?

Answer 1

Hint: First of all take all the terms to the L.H.S. Now, use the middle term split method to factorize the quadratic equation. Split 7y into two terms such that their sum equals 7y and the product equals \[12{{y}^{2}}\]. For this process, find the prime factors of 12 and combine them in a suitable way such that the conditions are satisfied. Finally, write the equation as the product of two linear binomials and equate them with 0 to find the values of y.

Complete step by step solution:
Here we have been asked to solve the quadratic equation \[7y=-3{{y}^{2}}-4\]. Here we will use the middle term split method to get the two values of y.
Now, taking all the terms to the L.H.S we get,
\[\Rightarrow 3{{y}^{2}}+7y+4=0\]
First we need to factorize the above expression. Let us use the middle term split method for the factorization. In this case we have to split the middle term which is 7y into two terms such that their sum is 7y and the product is equal to the product of constant term (4) and \[3{{y}^{2}}\], i.e. \[12{{y}^{2}}\]. To do this, first we need to find all the prime factors of 12.
We can write \[12=2\times 2\times 3\] as the product of its primes. Now, we have to group these factors such that our conditions of the middle terms split method are satisfied. So we have,
(i) \[\left( 4y \right)+\left( 3y \right)=7y\]
(ii) \[\left( 4y \right)\times \left( 3y \right)=12{{y}^{2}}\]
Hence, both the conditions of the middle term split method are satisfied. So, the quadratic equation can be written as: -
\[\begin{align}
  & \Rightarrow 3{{y}^{2}}+3y+4y+4=0 \\
 & \Rightarrow 3y\left( y+1 \right)+4\left( y+1 \right)=0 \\
\end{align}\]
Taking (y + 1) common we get,
\[\Rightarrow \left( y+1 \right)\left( 3y+4 \right)=0\]
Substituting each term equal to 0 we get,
\[\Rightarrow \left( y+1 \right)=0\] or \[\left( 3y+4 \right)=0\]
\[\Rightarrow y=-1\] or \[y=\dfrac{-4}{3}\]
Hence, the above two values of y are the roots of the quadratic equation.

Note: You can also use the discriminant method to solve the question. What you have to do is, after taking all the terms to the L.H.S apply the quadratic formula given as $y=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ and simplify it to get the two roots as answer. Here, a is the coefficient of ${{y}^{2}}$, b is the coefficient of y and c is the constant term.