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dg12300

03.02.2021

Mathematics

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

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