Question

How‌ ‌do‌ ‌you‌ ‌solve‌ ‌$3{{y}^{2}}+5y-4=0$‌ ‌using‌ ‌the‌ ‌quadratic‌ ‌formula?‌

Answer 1

Hint: The equation given in the above question is to be solved using the quadratic formula, which is given by $y=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$, where $a$, $b$ and $c$ are respectively the coefficients of ${{y}^{2}}$, $y$ and the constant term. From the given equation $3{{y}^{2}}+5y-4=0$ we note the coefficient of ${{y}^{2}}$ as $a=3$, the coefficient of $y$ as $b=5$ and the constant term as $c=-4$. On substituting these values of the coefficients into the quadratic formula, we will obtain the solutions of the given equation.

Complete step-by-step solution:
The given equation is
$\Rightarrow 3{{y}^{2}}+5y-4=0........(i)$
As we can observe from the above equation, it is in the form of the variable $y$ whose degree is equal to two. Therefore, the given equation is a quadratic equation in $y$ and so it will have two solutions.
In the above question, we are directed to solve the given equation using the quadratic formula, which is given by
$\Rightarrow y=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
From the given equation (i) we note the values of the coefficients as $a=3$, $b=5$, and $c=-4$. Substituting these into the quadratic formula written above, we get
\[\begin{align}
  & \Rightarrow y=\dfrac{-\left( 5 \right)\pm \sqrt{{{\left( 5 \right)}^{2}}-4\left( 3 \right)\left( -4 \right)}}{2\left( 3 \right)} \\
 & \Rightarrow y=\dfrac{-5\pm \sqrt{25-\left( -48 \right)}}{6} \\
 & \Rightarrow y=\dfrac{-5\pm \sqrt{73}}{6} \\
 & \Rightarrow y=\dfrac{-5+\sqrt{73}}{6},y=\dfrac{-5-\sqrt{73}}{6} \\
\end{align}\]
Hence, the solutions of the given equation are \[y=\dfrac{-5+\sqrt{73}}{6}\] and \[y=\dfrac{-5-\sqrt{73}}{6}\].

Note: Do not forget the negative sign in front of $b$ in the quadratic formula $y=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. Also, we must substitute the values of the coefficients with the proper signs from the given quadratic equation $3{{y}^{2}}+5y-4=0$. We can note that the constant term $c$ is negative and is equal to $-4$. So we must put it as it is and not put its absolute value. And the coefficient of ${{y}^{2}}$ is generally equal to one, but not in all the cases. Like in this case, it is equal to $3$, and so we must carefully put it in the quadratic formula.