Question

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

Answer 1

Answer:

$x=\frac{50+\sqrt{31}}{2},\frac{50-\sqrt{31}}{2}$  are zeroes of given quadratic equation.

Step-by-step explanation:

We have been a quadratic equation:

$2x^2-10x-3$

We need to find the zeroes of quadratic equation

We have a formula to find zeroes of a quadratic equation:

$x=\frac{b^2\pm\sqrt{D}}{2a}\text{where}D=\sqrt{b^2-4ac}$

General form of quadratic equation is $ax^2+bx+c$

On comparing general equation with b given equation we get

a=2,b=-10,c=-3

On substituting the values in formula we get

$D=\sqrt{(-10)^2-4(2)(-3)}$

$\Rightarrow D=\sqrt{100+24}=\sqrt{124}$

Now substituting D in  $x=\frac{b^2\pm\sqrt{D}}{2a}$ we get

$x=\frac{(-10)^2\pm\sqrt{124}}{2\cdot 2}$

$x=\frac{100\pm\sqrt{124}}{4}$

$x=\frac{100\pm2\sqrt{31}}{4}$

$x=\frac{50\pm\sqrt{31}}{2}$

Therefore, $x=\frac{50+\sqrt{31}}{2},\frac{50-\sqrt{31}}{2}$

Answer 2

Answer:

A quadratic equation is in the form of $ax^2+bx+c = 0$ .......[1]

then the solution for this equation is given by:

$x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$

Given the quadratic equation:

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

To find the zero of the given equation.

Set f(x) = 0

then;

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

On comparing with equation [1] we have;

a = 2, b = -10 and c = -3

then;

$x = \frac{-(-10) \pm \sqrt{(-10)^2-4(2)(-3)}}{2(2)}$

⇒$x = \frac{10 \pm \sqrt{100+24}}{4}$

⇒$x = \frac{10 \pm \sqrt{124}}{4}$

⇒$x = \frac{10 \pm 2\sqrt{31}}{4}$

Simplify:

$x = \frac{5 \pm \sqrt{31}}{2}$

Therefore, the zeros of the given quadratic equation are;

$x = \frac{5+\sqrt{31}}{2}$ and $x=\frac{5-\sqrt{31}}{2}$