
How do you solve the given quadratic equation $3{{x}^{2}}+7x+3=0$?
Answer
558k+ views
Hint: We start solving the problem by recalling the fact that the roots of the quadratic equation $a{{x}^{2}}+bx+c=0$ are defined as $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$, which is also known as quadratic formula. We then compare the given quadratic equation $3{{x}^{2}}+7x+3=0$ with $a{{x}^{2}}+bx+c=0$ to get the values of a, b and c. We then substitute these values in the quadratic formula and then make the necessary calculations to get the solution of the given quadratic equation.
Complete step-by-step solution:
According to the problem, we are asked to solve the given quadratic equation $3{{x}^{2}}+7x+3=0$.
We have given the quadratic equation $3{{x}^{2}}+7x+3=0$ ---(1).
We know that the roots of the quadratic equation $a{{x}^{2}}+bx+c=0$ are defined as $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$, which is also known as quadratic formula.
Now, let us compare the given quadratic equation $3{{x}^{2}}+7x+3=0$ with $a{{x}^{2}}+bx+c=0$. So, we get $a=3$, $b=7$, $c=3$. Now, let us substitute these values in quadratic formula to get the roots of the given quadratic equation $3{{x}^{2}}+7x+3=0$.
So, the roots are $x=\dfrac{-7\pm \sqrt{{{7}^{2}}-4\left( 3 \right)\left( 3 \right)}}{2\left( 3 \right)}$.
$\Rightarrow x=\dfrac{-7\pm \sqrt{49-36}}{6}$.
$\Rightarrow x=\dfrac{-7\pm \sqrt{13}}{6}$.
$\Rightarrow x=\dfrac{-7+\sqrt{13}}{6}$, $\dfrac{-7-\sqrt{13}}{6}$.
So, we have found the solution (roots) of the given quadratic equation $3{{x}^{2}}+7x+3=0$ as $\dfrac{-7+\sqrt{13}}{6}$, $\dfrac{-7-\sqrt{13}}{6}$.
$\therefore $ The solution (roots) of the given quadratic equation $3{{x}^{2}}+7x+3=0$ is $\dfrac{-7+\sqrt{13}}{6}$, $\dfrac{-7-\sqrt{13}}{6}$.
Note: We should perform each step carefully in order to avoid confusion and calculation mistakes. Whenever we get this type of problem, we first compare the given quadratic equation with the standard quadratic equation to get the required answer. We can also solve the given problem by factoring the given quadratic equation and then equating the factors to 0 to get the required roots. Similarly, we can expect problems to find the roots of the given quadratic equation $7{{x}^{2}}+6x-1=0$ using quadratic formula.
Complete step-by-step solution:
According to the problem, we are asked to solve the given quadratic equation $3{{x}^{2}}+7x+3=0$.
We have given the quadratic equation $3{{x}^{2}}+7x+3=0$ ---(1).
We know that the roots of the quadratic equation $a{{x}^{2}}+bx+c=0$ are defined as $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$, which is also known as quadratic formula.
Now, let us compare the given quadratic equation $3{{x}^{2}}+7x+3=0$ with $a{{x}^{2}}+bx+c=0$. So, we get $a=3$, $b=7$, $c=3$. Now, let us substitute these values in quadratic formula to get the roots of the given quadratic equation $3{{x}^{2}}+7x+3=0$.
So, the roots are $x=\dfrac{-7\pm \sqrt{{{7}^{2}}-4\left( 3 \right)\left( 3 \right)}}{2\left( 3 \right)}$.
$\Rightarrow x=\dfrac{-7\pm \sqrt{49-36}}{6}$.
$\Rightarrow x=\dfrac{-7\pm \sqrt{13}}{6}$.
$\Rightarrow x=\dfrac{-7+\sqrt{13}}{6}$, $\dfrac{-7-\sqrt{13}}{6}$.
So, we have found the solution (roots) of the given quadratic equation $3{{x}^{2}}+7x+3=0$ as $\dfrac{-7+\sqrt{13}}{6}$, $\dfrac{-7-\sqrt{13}}{6}$.
$\therefore $ The solution (roots) of the given quadratic equation $3{{x}^{2}}+7x+3=0$ is $\dfrac{-7+\sqrt{13}}{6}$, $\dfrac{-7-\sqrt{13}}{6}$.
Note: We should perform each step carefully in order to avoid confusion and calculation mistakes. Whenever we get this type of problem, we first compare the given quadratic equation with the standard quadratic equation to get the required answer. We can also solve the given problem by factoring the given quadratic equation and then equating the factors to 0 to get the required roots. Similarly, we can expect problems to find the roots of the given quadratic equation $7{{x}^{2}}+6x-1=0$ using quadratic formula.
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