Respuesta :
Answer:
A positive discriminant shows that the quadratic equation [tex]6x^2+13x+6=0[/tex] has two distinct real number solutions namely [tex]x_1=\frac{-2}{3}[/tex] or [tex]x_2=\frac{3}{2}[/tex]
Step-by-step explanation:
Consider the given Quadratic equation [tex]6x^2+13x+6=0[/tex]
Discriminant is a part of quadratic formula that shows the nature of root for a given quadratic equation.
Quadratic formula for a general quadratic equation of the form [tex]ax^2+bx+c=0[/tex] is given as:
[tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex] ......(A)
Where, [tex]\sqrt{b^2-4ac}[/tex] is the discriminant. .......(1)
The discriminant can be positive, zero, or negative.
A positive discriminant shows that the quadratic equation has two distinct real number solutions.
A discriminant of zero shows that the quadratic equation has a repeated real number solution.
A negative discriminant shows that neither of the solutions are real numbers.
Given Quadratic equation [tex]6x^2+13x+6=0[/tex]
Here, a= 6 , b=13, c= 6
Put in (1) , [tex]\sqrt{b^2-4ac}[/tex]
[tex]\Rightarrow \sqrt{(13)^2-4 \times 6 \times 6}[/tex]
[tex]\Rightarrow \sqrt{169-144}[/tex]
[tex]\Rightarrow \sqrt{25}[/tex]
[tex]\Rightarrow 5[/tex]
Thus, a positive discriminant shows that the quadratic equation has two distinct real number solutions.
Roots can be find as, Using (A)
[tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
[tex]x=\frac{-13\pm 5}{2 \times 6}[/tex]
[tex]x_1=\frac{-13+ 5}{12}[/tex] or [tex]x_2=\frac{-13- 5}{12}[/tex]
[tex]x_1=\frac{-8}{12}[/tex] or [tex]x_2=\frac{18}{12}[/tex]
[tex]x_1=\frac{-2}{3}[/tex] or [tex]x_2=\frac{3}{2}[/tex]
Hence, the system has two distinct real roots.
