Question 43.
Given the equation:
[tex]12x^2-8x-9=0[/tex]
Let's solve using the quadratic formula.
Apply the quadratic formula:
[tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
Use the standard formula to find the values of a, b and c:
[tex]\begin{gathered} ax^2+bx+c=0 \\ 12x^2-8x-9=0 \end{gathered}[/tex]
Where:
a = 12
b = -8
c = -9
Input values into the quadratic formula and solve for x:
[tex]\begin{gathered} x=\frac{-(-8)\pm\sqrt{(-8)^2-4(12)(-9)}}{2(12)} \\ \\ x=\frac{8\pm\sqrt{64-(-432)}}{24} \\ \\ x=\frac{8\pm\sqrt{64+432}}{24} \\ \\ x=\frac{8\pm\sqrt{496}}{24} \end{gathered}[/tex]
Solving further:
[tex]\begin{gathered} x=\frac{8\pm\sqrt{31*16}}{24} \\ \\ x=\frac{8\pm4\sqrt{31}}{24} \end{gathered}[/tex]
Now, let's simplify:
[tex]\begin{gathered} x=\frac{8}{24}\pm\frac{4\sqrt{31}}{24} \\ \\ x=\frac{1}{3}\pm\frac{1\sqrt{31}}{6} \\ \\ x=\frac{1}{3}\pm\frac{\sqrt{31}}{6} \end{gathered}[/tex]
ANSWER:
[tex]x=\frac{1}{3}\pm\frac{\sqrt{31}}{6}[/tex]
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