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

Last updated date: 27th Feb 2024
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IVSAT 2024
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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 $3{{x}^{2}}+4x-2=0$.
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
$a=3$, $b=4$, $c=-2$.
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
$\Rightarrow x=\dfrac{-\left( 4 \right)\pm \sqrt{{{\left( 4 \right)}^{2}}-4\left( 3 \right)\left( -2 \right)}}{2\left( 3 \right)}$
We know that when we multiplied a negative sign with the positive sign, then we will get negative sign. Applying the above rule and simplifying the above equation, then we will get
  & \Rightarrow x=\dfrac{-4\pm \sqrt{16+24}}{6} \\
 & \Rightarrow x=\dfrac{-4\pm \sqrt{40}}{6} \\
In the above equation we have the value $\sqrt{40}$. We need to simplify this value to get the simplified result. We can write $40=10\times 4=10\times {{2}^{2}}$, then the value of $\sqrt{40}$ will be $\sqrt{40}=\sqrt{10\times {{2}^{2}}}=2\sqrt{10}$. Substituting this value in the above equation, then we will get
$\Rightarrow x=\dfrac{-4\pm 2\sqrt{10}}{6}$
Calculating each value individually, then we will get
  & \Rightarrow x=\dfrac{-4+2\sqrt{10}}{6}\text{ or }\dfrac{-4-2\sqrt{10}}{6} \\
 & \Rightarrow x=\dfrac{2\left( -2+\sqrt{10} \right)}{6}\text{ or }\dfrac{2\left( -2-\sqrt{10} \right)}{6} \\
 & \Rightarrow x=\dfrac{-2+\sqrt{10}}{3}\text{ or }\dfrac{-2-\sqrt{10}}{3} \\

Hence the solution of the given quadratic equation $3{{x}^{2}}+4x-2=0$ are $x=\dfrac{-2\pm \sqrt{10}}{3}$.

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 $3{{x}^{2}}+4x-2=0$ it looks like below graph
seo images

From the above graph also, we can say that the roots of the given equation $3{{x}^{2}}+4x-2=0$ are $x=\dfrac{-2\pm \sqrt{10}}{3}$.
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