Respuesta :
Given:
[tex]3x^2+2x=2[/tex]Question 2:
Let's find the values of a, b, and c.
Step 1:
Equate the equation to zero
Subtract 2 from both sides
[tex]\begin{gathered} 3x^2+2x-2=2-2 \\ \\ 3x^2+2x-2=0 \end{gathered}[/tex]To find the values of a, b, and c, apply the general quadratic formula:
[tex]ax^2+bx+c=0[/tex]Comapre both equations:
[tex]\begin{gathered} ax^2+bx+c=0 \\ 3x^2+2x-2=0 \end{gathered}[/tex]Thus, we have the following values:
a = 3
b = 2
c = -2
Question 3:
Let's solve for x
To solve for x, apply the formula:
[tex]x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}[/tex]
Substitute 3 for a, 2 for b, and -2 for c:
[tex]\begin{gathered} x=\frac{-2\pm\sqrt[]{2^2-4(3)(-2)}}{2(3)} \\ \\ x=\frac{-2\pm\sqrt[]{4-(-24)}}{6} \\ \\ x=\frac{-2\pm\sqrt[]{4+24}}{6} \\ \\ x=\frac{-2\pm\sqrt[]{28}}{6} \end{gathered}[/tex]Solving further:
[tex]\begin{gathered} x=\frac{-2\pm5.29}{6} \\ \\ x=\frac{-2+5.29}{6},\text{ }\frac{-2-5.29}{6} \\ \\ x=0.548,\text{ }-1.215 \end{gathered}[/tex]Therefore, the values of x are:
x = 0.548, -1.215
ANSWER:
2. a = 3, b = 2, c = -2
3. x = 0.548
x = -1.215
