Question

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

Answer 1

x= -0.4 , x= -2.6

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= (-b+-√(b^2-4ac))/(2a)`

a=1 , b= 3, c=1

`x= (-3+-√(3^2-4(1)(1)))/(2(1))`

`x= (-3+-√(5))/(2)`

LEts make two equations

`x= (-3+√(5))/(2)`

x= -0.4

`x= (-3-√(5))/(2)`

x= -2.6

Answer 2

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