Question

Approximate the real zeros of f(x)=x^2+3x+1 to the nearest tenth.

Answer 1

we have that

f(x)=x²+3x+1
using a graph tool
see the attached figure

the solutions are
x=-2.618
x=-0.382

the answer is
x=-2.6
x=-0.4

Answer 2

Answer:

x= -0.4  , x= -2.6

Step-by-step explanation:

f(x)=x^2+3x+1

to find out zeros of the given f(x) we replace f(x) with 0.

0= x^2 + 3x +1

Then we apply quadratic formula

$x= \frac{-b+-\sqrt{b^2-4ac}}{2a}$

a=1 , b= 3, c=1

$x= \frac{-3+-\sqrt{3^2-4(1)(1)}}{2(1)}$

$x= \frac{-3+-\sqrt{5}}{2}$

LEts make two equations

$x= \frac{-3+\sqrt{5}}{2}$  

x= -0.4  

$x= \frac{-3-\sqrt{5}}{2}$

x= -2.6