Question

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

Answer 1

`x=(50+√(31))/(2),(50-√(31))/(2)` are zeroes of given quadratic equation.

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=(b^2\pm√(D))/(2a)\text{where}D=√(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=√((-10)^2-4(2)(-3))`

`\Rightarrow D=√(100+24)=√(124)`

Now substituting D in
`x=(b^2\pm√(D))/(2a)` we get

`x=((-10)^2\pm√(124))/(2\cdot 2)`

`x=(100\pm√(124))/(4)`

`x=(100\pm2√(31))/(4)`

`x=(50\pm√(31))/(2)`

Therefore,
`x=(50+√(31))/(2),(50-√(31))/(2)`

Answer 2

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

then the solution for this equation is given by:

`x = (-b \pm √(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 = (-(-10) \pm √((-10)^2-4(2)(-3)))/(2(2))`

⇒
`x = (10 \pm √(100+24))/(4)`

⇒
`x = (10 \pm √(124))/(4)`

⇒
`x = (10 \pm 2√(31))/(4)`

Simplify:

`x = (5 \pm √(31))/(2)`

Therefore, the zeros of the given quadratic equation are;

`x = (5+√(31))/(2)` and
`x=(5-√(31))/(2)`