Homework Help: College Algebra--Still Confused

Posted by Krissy on Tuesday, February 5, 2013 at 11:18am.

I have a few problems I need help with and also do have multiple choice. If I can have an explanation of how to get the answer that would be great.

1. Use the discriminant to determine whether the given equation has two irrational roots, two rational roots, one repeated root, or two complex roots.
x^2-4x=5
a.) Two rational roots
b.) one repreated root
c.) two complex roots
d.) two irrational roots.

2.Use the discriminant to determine whether the given equation has two irrational roots, two rational roots, one repeated root, or two complex roots.
-3x^3+19x-20=0
a.) Two rational roots
b.) one repreated root
c.) two complex roots
d.) two irrational roots.

3. Use the discriminant to determine whether the given equation has two irrational roots, two rational roots, one repeated root, or two complex roots.
-5x^2-7x-5=0
a.) Two rational roots
b.) one repreated root
c.) two complex roots
d.) two irrational roots.

4. Use the discriminant to determine whether the given equation has two irrational roots, two rational roots, one repeated root, or two complex roots
x^2-12=7x
a.) Two rational roots
b.) one repreated root
c.) two complex roots
d.) two irrational roots.

College Algebra--Still Confused - Bosnian, Tuesday, February 5, 2013 at 11:23am
In google type:

quadratic equation online

When you see list of result click on:

Free Online Quadratic Equation Solver:Solve by Quadratic Formula

When page be open in rectangle type your equation

and cilck option:

solve it!

You will see solutions step-by step and discriminant discusion.
College Algebra--Still Confused - Damon, Tuesday, February 5, 2013 at 11:27am
I am only going to do one,

1. Use the discriminant to determine whether the given equation has two irrational roots, two rational roots, one repeated root, or two complex roots.
x^2-4x=5
a.) Two rational roots
b.) one repreated root
c.) two complex roots
d.) two irrational roots.
----------------
Write with a zero on the right in form
a x^2 + b x + c = 0
1 x^2 -4 x - 5 = 0
b^2 = (-4)(-4) = 16
-4ac = -4(1)(-5) = +20
so
b^2 -4ac = 16+20 = 36
sqrt (b^2 - 4ac) = +6
THAT means:
x = [ 4 +/- 6 ] /2
Two REAL Rational roots
BECAUSE
sqrt (+36) is a real number
sqrt (+36) is not sqrt( 0) which would give one repeated root
sqrt(+36) is not sqrt(-36) which would leave you with a +/- 6i which is COMPLEX number
it is not sqrt (2) or something which is not a nice rational number and can not be expressed as a ratio of whole numebers

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