MATH SOLVE

2 months ago

Q:
# in simplest radical form, what are the solutions to the quadratic eqaution 0=-3x^2-4x+5

Accepted Solution

A:

Answer:[tex]x= -\frac{2+\sqrt{19} }{3} \textrm { and } x = -\frac{2-\sqrt{19} }{3}[/tex] Step-by-step explanation:We have to find the solution of the quadratic single variable equation as given below:
- 3x² - 4x + 5 = 0 ..... (1)
The left-hand side can not be factorized.
So, apply The Sridhar Acharya Formula, which gives if ax² + bx + c = 0,the the solutions of the quadratic equation are
[tex]x=\frac{-b + \sqrt{b^{2} -4ac} }{2a}[/tex] and, [tex]x=\frac{-b - \sqrt{b^{2} -4ac} }{2a}[/tex]
So, from the equation (1), we can write [tex]x= \frac{-(-4) +\sqrt{(-4)^{2}-4 \times (-3) \times 5 } }{2 \times (-3)}[/tex] and
[tex]x= \frac{-(-4) -\sqrt{(-4)^{2}-4 \times (-3) \times 5 } }{2 \times (-3)}[/tex]
⇒ [tex]x=\frac{4+ 2\sqrt{19} }{-6} \textrm{ and } x=\frac{4-2\sqrt{19} }{-6}[/tex]
⇒ [tex]x= -\frac{2+\sqrt{19} }{3} \textrm { and } x = -\frac{2-\sqrt{19} }{3}[/tex] (Answer)