Question

How do you solve $2{{x}^{2}}+6x+4=0$ using the quadratic formula?

Answer 1

Hint: In this problem we need to solve the given quadratic equation i.e., we need to calculate the values of $x$ where the given equation is satisfied. For solving a quadratic equation, we have several methods. But in the problem, they have mentioned to use the quadratic formula which is given by $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. For this we need to compare the given equation with the standard quadratic equation $a{{x}^{2}}+bx+c=0$ and write the values of $a$, $b$, $c$. Now we will substitute those values in the formula $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ and simplify the obtained equation to get the required result.

Complete step by step answer:
Given equation $2{{x}^{2}}+6x+4=0$.
Comparing the above quadratic equation with standard quadratic equation $a{{x}^{2}}+bx+c=0$, then we will get the values of $a$, $b$, $c$ as
$a=2$, $b=6$, $c=4$.
We have the quadratic formula for the solution as
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Substituting the values of $a$, $b$, $c$ in the above equation, then we will get
$\Rightarrow x=\dfrac{-\left( 6 \right)\pm \sqrt{{{\left( 6 \right)}^{2}}-4\left( 2 \right)\left( 4 \right)}}{2\left( 2 \right)}$
We know that when we multiplied a negative sign with a positive sign, then we will get a negative sign. Applying the above rule and simplifying the above equation, then we will get
$\begin{align}
& \Rightarrow x=\dfrac{-6\pm \sqrt{36-32}}{4} \\
& \Rightarrow x=\dfrac{-6\pm \sqrt{4}}{4} \\
\end{align}$
In the above equation we have the value $\sqrt{4}$. We need to simplify this value to get the simplified result. We can write $4=2\times 2={{2}^{2}}$, then the value of $\sqrt{4}$ will be $\sqrt{4}=\sqrt{{{2}^{2}}}=2$. Substituting this value in the above equation, then we will get
$\Rightarrow x=\dfrac{-6\pm 2}{4}$
Taking $2$ as common in the numerator, then we will get
$\begin{align}
& \Rightarrow x=\dfrac{2\left( -3\pm 1 \right)}{4} \\
& \Rightarrow x=\dfrac{-3\pm 1}{2} \\
\end{align}$
Calculating each value individually, then we will get
$\begin{align}
& \Rightarrow x=\dfrac{-3+1}{2}\text{ or }\dfrac{-3-1}{2} \\
& \Rightarrow x=\dfrac{-2}{2}\text{ or }\dfrac{-4}{2} \\
& \Rightarrow x=-1\text{ or }-2 \\
\end{align}$

Hence the solution of the given quadratic equation $2{{x}^{2}}+6x+4=0$ are $x=-1,-2$.

Note: We can also see the graph of the above given equation to observe the roots of the equation. When we plot the graph of the given equation $2{{x}^{2}}+6x+4=0$ it looks like a parabola intersecting the x-axis twice.