The zeroes of the function f(x) = x² + 8x + 4 is calculated by shri dharachaya formula negative 0.54 and negative 7.46.
The function is an expression, rule, or law that defines the relationship between one variable to another variable. Functions are ubiquitous in mathematics and are essential for formulating physical relationships.
The function f(x) = x² + 8x + 4
The zeroes of the function will be given as
f(x) = 0
Then we have
x² + 8x + 4 = 0
a = 1, b = 8, c = 4
By the formula
[tex]\rm x = \dfrac{-b \pm \sqrt{b^2 - 4ac }}{2a}[/tex]
Then
[tex]\rm x = \dfrac{-8 \pm \sqrt{8^2 - 4 * 1 *4}}{2*1}\\\\x = \dfrac{-8 \pm \sqrt{64 - 16}}{2}\\\\x = \dfrac{-8 \pm 4 \sqrt{3}}{2}\\\\x = -4 \pm 2\sqrt3\\\\x = -0.54, -7.46[/tex]
More about the function link is given below.
https://brainly.com/question/5245372
Answer:
x= -4 +/- 2 [tex]\sqrt{3}[/tex]
Step-by-step explanation:
