Question

What are the zeros of the quadratic function f(x) = 2x2 – 10x – 3?

Answer 1

When 2x^2 - 10x - 3 is plugged into the quadratic equation, the resulting zeroes are x = (5 + sqrt(31))/2 and (5 - sqrt(31))/2.

Hope this helps!

Answer 2

we have

$f(x)=2x^{2}-10x-3$

we know that

To find the zeros of the quadratic equation equate to zero

$2x^{2}-10x-3=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$2x^{2}-10x=3$

Factor the leading coefficient

$2(x^{2}-5x)=3$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$2(x^{2}-5x+6.25)=3+12.5$

$2(x^{2}-5x+6.25)=15.5$

Rewrite as perfect squares

$2(x-2.5)^{2}=15.5$

$(x-2.5)^{2}=7.75$

$(x-2.5)=(+/-)\sqrt{7.75}$

$(x-2.5)=(+/-)2.78$

$x1=(+)2.78+2.5=5.284$  or  $x1=\frac{\sqrt{31}}{2} +2.5$

$x2=(-)2.78+2.5=-0.284$  or  $x2=-\frac{\sqrt{31}}{2} +2.5$

therefore

the answer is

$x1=5.284$  or  $x1=\frac{\sqrt{31}}{2} +2.5$

$x2=-0.284$  or   $x2=-\frac{\sqrt{31}}{2} +2.5$

see the attached figure to verify