Answer:
[tex]x=\frac{1+\sqrt{35}i}{-6}\,\, and\,\, x=\frac{1-\sqrt{35}i}{-6}\\[/tex]
Step-by-step explanation:
the quadratic formula is:
[tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
a= -2, b = -1 and c =-3
Putting values in the formula
[tex]x=\frac{-(-1)\pm\sqrt{(-1)^2-4(-3)(-3)}}{2(-3)}\\x=\frac{1\pm\sqrt{-35}}{-6}\\x=\frac{1+\sqrt{-35}}{-6}\,\, and\,\, x=\frac{1-\sqrt{-35}}{-6}\\We\,\, know \,\,that \,\,\sqrt{-1} = i \\x=\frac{1+\sqrt{35}i}{-6}\,\, and\,\, x=\frac{1-\sqrt{35}i}{-6}\\[/tex]
So, [tex]x=\frac{1+\sqrt{35}i}{-6}\,\, and\,\, x=\frac{1-\sqrt{35}i}{-6}\\[/tex]
Answer:
Using quadratic formula, the solution to this equation is the roots of the equations given are ; x = 1+√35i / -6 or x = 1-√35i / -6
Step-by-step explanation:
-3x² - x - 3=0
To solve this using quadratic formula, we will first of all write down the quadratic formula
x = -b ±√b²- 4ac / 2a
From the above question;
a = -3 b = -1 and c=-3
So we can now proceed to plug-in our variable
x = -(-1) ± √(-1)² - 4(-3)(-3) / 2(-3)
x= 1±√1-36 / -6
x = 1 ±√-35 / -6
x=1 ± √35 · √-1 /-6
x = 1±√35 i / -6
Note the square root of negative 1 is i
Either x = 1+√35i / -6 or x = 1-√35i / -6
Therefore the roots of the equations given are ; x = 1+√35i / -6 or x = 1-√35i / -6
