Respuesta :
[tex]f(b)=b^2-75
\\f(b)=0
\\b^2-75=0
\\b^2=75
\\b=\pm\sqrt{75}=\pm\sqrt{25\times3}=\pm5\sqrt{3}[/tex]
Answer:
The roots of the given quadratic function [tex]f(b) = b^2-75[/tex] is [tex]5\sqrt{3}\quad[/tex] and [tex]-5\sqrt{3}\quad[/tex]
Step-by-step explanation:
Given: Quadratic function [tex]f(b) = b^2-75[/tex]
We have to find the roots of the given quadratic function [tex]f(b) = b^2-75[/tex]
Since, roots of the quadratic equation is the points where the value of function is zero.
That is f(x) = 0
Consider the given function [tex]f(b) = b^2-75[/tex]
Put f(b) = 0
[tex]\Rightarrow b^2-75=0[/tex]
Simplify , we have,
[tex]\Rightarrow b^2=75[/tex]
Taking square root both side, we have,
[tex]\Rightarrow b=\sqrt{75}[/tex]
Simplify we have,
[tex]\Rightarrow b=\pm 5\sqrt{3}\quad[/tex]
Thus, The roots of the given quadratic function [tex]f(b) = b^2-75[/tex] is [tex]5\sqrt{3}\quad[/tex] and [tex]-5\sqrt{3}\quad[/tex]