Question

Solve:\[3{{x}^{2}}+13x+10=0\]

Answer 1

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.

Complete step by step answer:
 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.