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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)

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carlosego
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.

A) Consider the equation [tex]10x^2-3x-1=0[/tex]

We have to describe the solutions to this equation.

We will determine the radicand = [tex]b^2-4ac[/tex]

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

Therefore, the given equation ha real roots.

B) Consider the equation [tex]16x^2-2x-5=0[/tex]

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

[tex]16x^2+8x-10x-5 =0[/tex]

[tex]8x(2x+1) -5(2x+1)=0[/tex]

[tex](2x+1)(8x-5)=0[/tex]

So, [tex]2x+1=0[/tex]

[tex]x = \frac{-1}{2}[/tex] and [tex]x = \frac{5}{8}[/tex]

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

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