Question

A quadratic equation is shown below: 10x2 − 3x − 1 = 0 Part A: Describe the solution(s) to the equation by just determining the radicand. Show your work. (5 points) Part B: Solve 16x2 − 2x − 5 = 0 by using an appropriate method. Show the steps of your work, and explain why you chose the method used. (5 points)

Answer 1

A) Consider the equation
`10x^2-3x-1=0`

We have to describe the solutions to this equation.

We will determine the radicand =
`b^2-4ac`

=
`(-3)^2 - 4(10)(-1) = 9+ 40 = 49` which is greater than zero,

Therefore, the given equation ha real roots.

B) Consider the equation
`16x^2-2x-5=0`

We will use middle splitting term method to solve this equation.

`16x^2+8x-10x-5 =0`

`8x(2x+1) -5(2x+1)=0`

`(2x+1)(8x-5)=0`

So,
`2x+1=0`

`x = (-1)/(2)` and
`x = (5)/(8)`

I chose this method, because this is more easier and takes less time.

Answer 2

Part A:
For this case we have the following equation:
10x2 - 3x - 1 = 0
The radicand of the equation is:
root (b ^ 2 - 4 * a * c)
Substituting values:
root ((- 3) ^ 2 - 4 * (10) * (- 1))
root (9 + 40)
root (49)
49> 0
Therefore, the function has two real roots.

Part B:
For this case we have the following equation:
16x2 - 2x - 5 = 0
Factoring we have:
(2x + 1) * (8x-5) = 0
The solutions are:
x = -1/2
x = 5/8
The method of factoring is usually faster when solving quadratic equations.