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# Solve:$3{{x}^{2}}+13x+10=0$

Last updated date: 24th Jul 2024
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Hint: Compare the given equation with the general quadratic equation $a{{x}^{2}}+bx+c=0$ to obtain the values of a, b, and c. substitute these values in the quadratic equation formula $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ to compare the required roots.

We are given a quadratic equation $3{{x}^{2}}+13x+10=0$
How do we know this is a quadratic equation? Well, any equation of the form $a{{x}^{2}}+bx+c=0$ where a, b, and c are real numbers, and a is a non-zero constant is called a quadratic equation.
Now, the values of x which will satisfy the quadratic equation will be called the roots of the quadratic equation. Also, a quadratic equation will always have two roots. The roots may be the same or different.
Thus, we are going to locate the two roots of the given quadratic equation $3{{x}^{2}}+13x+10=0$.
To do this, we may use the quadratic equation formula.
The formula to finding the two roots of the quadratic equation $a{{x}^{2}}+bx+c=0$ is as follows:
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Comparing the given equation $3{{x}^{2}}+13x+10=0$ with the general quadratic equation, we have
a=3, b=13, and c=10.
Now, substitute these values in the quadratic equation formula.
This will give us
$x=\dfrac{-13\pm \sqrt{{{13}^{2}}-4\times 3\times 10}}{2\times 3}$
$\Rightarrow x=\dfrac{-13\pm \sqrt{169-120}}{6}$
On further solving,
$\Rightarrow x=\dfrac{-13\pm \sqrt{49}}{6}$
$\Rightarrow x=\dfrac{-13\pm 7}{6}$
Thus, we have $x=\dfrac{-13+7}{6}$ or $x=\dfrac{-13-7}{6}$
$\Rightarrow x=\dfrac{-6}{6}$ or $x=\dfrac{-20}{6}$
Reducing the fractions in their lowest form, we get $x=-1$ or $x=\dfrac{-10}{3}$.
Therefore, $\dfrac{-10}{3}$ and $-1$ are the roots of the equation $3{{x}^{2}}+13x+10=0$

Note: Alternate method to compute the roots of a quadratic equation is factorization method. Generally, the discriminant determines the nature of roots of a quadratic equation. The word ‘nature’ refers to the types of number the roots can be namely real, rational, irrational or imaginary.
If the discriminant $\sqrt{{{b}^{2}}-4ac}=0$, then the roots are equal and real
If the discriminant$\sqrt{{{b}^{2}}-4ac}>0$, then the roots are real
If the discriminant$\sqrt{{{b}^{2}}-4ac}<0$, then the roots are imaginary.