Question

What are the zeros of the quadratic function f(x)=2x^2-10x-3

Answer 1

5/2+(1/2)*sqrt(31), 5/2-(1/2)*sqrt(31)

Answer 2

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 $x=\dfrac{5-\sqrt{31}}{2}$