SOLUTION: How do you find if an equation has 2 real number solutions, 1 real solution or no real-number solutuons? an example is x^2-6x+c=0

Click here to see ALL problems on Quadratic Equations

Question 116197: How do you find if an equation has 2 real number solutions, 1 real solution or no real-number solutuons? an example is x^2-6x+c=0
Answer by jim_thompson5910(35256) (Show Source):
You can put this solution on YOUR website!
From the quadratic formula

the discriminant consists of all of the terms in the square root. So the discriminant is

the discriminant tells us how many solutions (and what type of solutions) we can expect for any quadratic.

So let's find the discriminant for

Plug in a=1, b=-6

Square -6 to get 36

Multiply -4*1*c to get -4c

------------------

Now if the discriminant is greater than zero, then we'll have 2 real solutions.

Set the discriminant greater than zero

Subtract 36 from both sides

Combine like terms on the right side

Divide both sides by -4 to isolate c (note: Remember, dividing both sides by a negative number flips the inequality sign)

Divide

So when , we'll have two real solutions.

Now when the discriminant is equal to zero, then we'll have one real solution:

Set the discriminant equal to zero

Subtract 36 from both sides

Combine like terms on the right side

Divide both sides by -4 to isolate c

Divide

So when , we'll have one real solution.

Now when the discriminant is less than zero, then we'll have no real solutions:

Set the discriminant less than zero

Subtract 36 from both sides

Combine like terms on the right side

Divide both sides by -4 to isolate c (note: Remember, dividing both sides by a negative number flips the inequality sign)

Divide

So when , we won't have any real solutions