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]

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