Really really need 3consider the quadratic equation x2 – 6x = –1. a. what is the value of the discriminant? explain. b. how many solutions does the quadratic equation have and are those solutions rational, irrational, or nonreal? explain. c. if the quadratic equation has real solutions, what are the solutions? explain. estimate irrational solutions to the nearest tenth.

Part A

Get everything to one side. The quickest way to do this is to add 1 to both sides
x^2 - 6x = -1
x^2 - 6x + 1 = -1+1
x^2 - 6x + 1 = 0

The equation is now in the form ax^2 + bx + c = 0
We see that,
a = 1
b = -6
c = 1

Plug those values into the discriminant formula
d = b^2 - 4ac
d = (-6)^2 - 4*1*1
d = 36 - 4
d = 32

The value of the discriminant is 32
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Part B

In part A, we got d = 32 as the discriminant. The discriminant is positive (greater than 0) so there are 2 different real number solutions. These solutions are irrational because d = 32 is not a perfect square. Taking the square root of 32 leads to a decimal value that goes on forever without any known pattern. It is impossible to represent sqrt(32) as a fraction of two whole numbers.

In short, the two solutions are irrational. 
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Part C

Use the quadratic formula to solve for x. Check out the image attachment for more detailed steps. The two irrational solutions, rounded to the nearest tenth, are x = 0.2 or x = 5.8

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