Answer:
The zeros of the quadratic function \(f(x)=2x^2-10x-3\) are:
\(x=\dfrac{5+\sqrt{31}}{2}\ ,\ x=\dfrac{5-\sqrt{31}}{2}\)
Step-by-step explanation:
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:
\(f(x)=2x^2-10x-3\)
Now, f(x)=0
\(2x^2-10x-3=0\)
We know that the solution of the quadratic equation of the type:
\(ax^2+bx+c=0\) is given by:
\(x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}\)
Here we have:
a=2, b= -10 and c= -3
Hence, the solution is:
\(x=\dfrac{-(-10)\pm \sqrt{(-10)^2-4\times 2\times (-3)}}{2\times 2}\\\\\\x=\dfrac{10\pm \sqrt{100+24}}{4}\\\\\\x=\dfrac{10\pm \sqrt{124}}{4}\\\\\\x=\dfrac{10\pm 2\sqrt{31}}{4}\\\\\\x=\dfrac{5\pm \sqrt{31}}{2}\)
Hence,
\(x=\dfrac{5+\sqrt{31}}{2}\) and \(x=\dfrac{5-\sqrt{31}}{2}\)
