Answer:
Using the quadratic formula
[tex]x=\frac{-b+/-\sqrt{b^{2}-4ac } }{2a}[/tex]
The answer to the equation [tex]2x^{2} +31x-6.1=0[/tex] using at least three significant figures is:
[tex]x_{1}=0.194\\x_{2}=-15.694[/tex]
Step-by-step explanation:
The quadratic formula is used to solve polynomials of second degree.
We have a polynomial of second degree to be resolved with the quadratic formula:
[tex]2x^{2} +31x-6.1=0[/tex] (Eq. 1)
We know the quadratic formula is:
[tex]x=\frac{-b+/-\sqrt{b^{2}-4ac } }{2a}[/tex] (Eq. 2)
To resolve the quadratic formula we need the a, b and c coefficients, we can find these coefficients in the equation 1.
a: Coefficient that accompanies [tex]x^{2}[/tex]
b: Coefficient that accompanies [tex]x[/tex]
c: Independent term
With this information and the equation (1). We know the values of a, b and c
[tex]a=2\\b=31\\c=-6.1\\[/tex]
Now, we can replace these terms in the quadratic formula (Eq. 2)
The first root will be found using the positive sign before the square root:
[tex]x=\frac{-b+\sqrt{b^{2}-4ac } }{2a}[/tex]
[tex]x=\frac{-31+\sqrt{31^{2}-[4*2*(-6.1)]} }{2*2}[/tex]
[tex]x=\frac{-31+\sqrt{961-(-48.8)} }{4}[/tex]
[tex]x=\frac{-31+\sqrt{961+48.8} }{4}[/tex]
[tex]x=\frac{-31+\sqrt{1009.8} }{4}[/tex]
[tex]x=\frac{-31+31.777 }{4}[/tex]
[tex]x=0.194[/tex]
The second root will be found using the negative sign before the square root
:
[tex]x=\frac{-b-\sqrt{b^{2}-4ac } }{2a}[/tex]
[tex]x=\frac{-31-\sqrt{31^{2}-[4*2*(-6.1)]} }{2*2}[/tex]
[tex]x=\frac{-31-\sqrt{961-(-48.8)} }{4}[/tex]
[tex]x=\frac{-31-\sqrt{961+48.8} }{4}[/tex]
[tex]x=\frac{-31-\sqrt{1009.8} }{4}[/tex]
[tex]x=\frac{-31-31.777 }{4}[/tex]
[tex]x=-15.694[/tex]
