The zeros of the quadratic function [tex]f(x)=2x^2-10x-3[/tex] are:
[tex]x=\dfrac{5+\sqrt{31}}{2}\ ,\ x=\dfrac{5-\sqrt{31}}{2} [/tex]
Zeros of a function are the possible x values of the function for which the function is equal to zero.
i.e. all those x for which f(x)=0
We are given a function f(x) by:
[tex]f(x)=2x^2-10x-3[/tex]
Now, f(x)=0
[tex]2x^2-10x-3=0[/tex]
We know that the solution of the quadratic equation of the type:
[tex]ax^2+bx+c=0[/tex] is given by:
[tex]x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]
Here we have:
a=2, b= -10 and c= -3
Hence, the solution is:
[tex]x=\dfrac{-(-10)\pm \sqrt{(-10)^2-4\times 2\times (-3)}}{2\times 2}\\\\\\x=\dfrac{10\pm \sqrt{100+24}}{4}\\\\\\x=\dfrac{10\pm \sqrt{124}}{4}\\\\\\x=\dfrac{10\pm 2\sqrt{31}}{4}\\\\\\x=\dfrac{5\pm \sqrt{31}}{2}[/tex]
Hence,
[tex]x=\dfrac{5+\sqrt{31}}{2}[/tex] and [tex]x=\dfrac{5-\sqrt{31}}{2}[/tex]
