The roots are:
[tex]x1 = \frac{-3 + \sqrt{ 29 } }{2}[/tex]
and
[tex]x2 = \frac{-3 - \sqrt{ 29 } }{2}[/tex]
Step-by-step explanation:
We have to find the values of x that are roots of the polynomial x^2+3x-5.
For that, we can use quadratic equation, that is
[tex]x = \frac{-b \± \sqrt{b^{2} - 4ac } }{2a}[/tex]
The above equation is already in standard form ([tex]ax^{2} + bx + c = 0[/tex])
So,
a = 1
b = 3
c = -5
Putting the values in equation, we get
[tex]x = \frac{-3 \± \sqrt{ 3^{2} - 4(1)(-5) } }{2(1)}[/tex]
[tex]x = \frac{-3 \± \sqrt{ 9 + 20 } }{2}[/tex]
[tex]x = \frac{-3 \± \sqrt{ 29 } }{2}[/tex]
Therefore, the roots are:
[tex]x1 = \frac{-3 + \sqrt{ 29 } }{2}[/tex]
and
[tex]x2 = \frac{-3 - \sqrt{ 29 } }{2}[/tex]
Keywords: roots, polynomial
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Answer:
It is -3-square root of 11/2 and -3+square root of -11/2
Step-by-step explanation:
