Please Help With Math Homework!?

Question

I'm doing an algerbra review and I don't remember how you can tell if an equation has one solution, two solutions or no solution... Could anyone explain how?

Some problems are...
5x^2-9x+16=0
x^2=2x-1
and
3x^2-x-1=0

Answer 1

You would use what's called the "discriminant" which is b^2-4ac when the equation is in the form of ax^2 +bx +c =0.

If b^2 -4ac > 0, there are two solutions
If it =0 , there is one solution
if it <0 , there are no solutions

Answer 2

Use the quadratic formula, which is x = [-b +/- sqrt(b^2-4ac)]/2a. In the case of 5x^2-9x+16=0 a=5, b=(-9), and c=16, so we have [9 +/- sqrt(81-4(5)(16)]/2(5). Since 81-320 is a negative number, you cannot take the sqrt(-239), so in this case there is no solution.

For the second equation above, you must put it into the form ax^2+bx+c=0, so convert it to x^2-2x+1=0. Then use the quadratic formula. We have [2+/- sqrt(4-4)]/2], which is 2/2 = 1. x=1 is the only solution because 2-sqrt(0) and 2+sqrt(0) is the same thing.

For the third equation we have [1+/- sqrt(1-4(3)(-1)]/2(3), or [1-sqrt(13)]/6 and [1+sqrt(13)]/6 -- two solutions.

Visit the following site: http://www.purplemath.com/modules/quadform.htm

It'll give you additional info on the quadratic formula.

Answer 3

graph it with your graphing calculator and see how many times it crosses the x axis