SOLUTION: How do I find the discriminant? The question ask to find {{{b^2-4ac}}} and the number of real solutions to each equation. 35. {{{4m^2+25 = 20m}}} 45. {{{x^2=x}}}

Algebra ->  Radicals -> SOLUTION: How do I find the discriminant? The question ask to find {{{b^2-4ac}}} and the number of real solutions to each equation. 35. {{{4m^2+25 = 20m}}} 45. {{{x^2=x}}}      Log On


   



Question 157288This question is from textbook elementary and intermediate algebra
: How do I find the discriminant? The question ask to find b%5E2-4ac and the number of real solutions to each equation.
35. 4m%5E2%2B25+=+20m
45. x%5E2=x
This question is from textbook elementary and intermediate algebra

Found 2 solutions by Edwin McCravy, SAT Math Tutor:
Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
How do I find the discriminant? The question ask to find b%5E2-4ac and the number of real solutions to each equation.
35. 4m%5E2%2B25+=+20m
45. x%5E2=x

You must first rearrange the equations so that 0 appears
on the right and the three terms on the left are in 
descending order like this:

am%5E2+%2B+bm+%2B+c+=+0, ax%5E2%2Bbx%2Bc=0 or whatever 
the letter of the unknown happens to be:

That's the only way you can tell what to substitute for
a, b, and c in the formula:

DISCRIMINANT=b%5E2-4ac

-------

For your problem 35:

4m%5E2%2B25+=+20m

You must first get a 0 on the right side by adding -20m to
both sides:

4m%5E2%2B25-20m=20m-20m

4m%5E2%2B25-20m=0

And you must write the left side in descending order, like 
this:

4m%5E2-20m%2B25=0

Then when you compare that to

am%5E2+%2B+bm+%2B+c+=+0, 

it is easy to see that a=4,b=-20, and c=25.

So then you can easily substitute those values in the 
expression for the discriminant:

DISCRIMINANT=b%5E2-4ac

DISCRIMINANT=%28-20%29%5E2-4%284%29%2825%29

DISCRIMINANT=400-400

DISCRIMINANT=0

When the discriminant is 0, there is 
exactly 1 real solution. So this particular
quadratic equation has 1 real solution.


If it were negative there would be no real solutions,
and if it were positive there would be exactly two real
solutions.

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

For your problem 45:

x%5E2=x

You must first get a 0 on the right side by adding -x to
both sides:

x%5E2-x=x-x

x%5E2-x=0

But the left side contains only two terms! So you must 
add on a 0 to the left side:

x%5E2-x%2B0=0

Also it may be helpful to write the understood 1's before
x%5E2 and x, like this:

1x%5E2-1x%2B0=0

The left side is already in descending order, so when you 
compare that to

ax%5E2+%2B+bx+%2B+c+=+0, 

it is easy to see that a=1,b=-1, and c=0.

So then you can easily substitute those values in the 
expression for the discriminant:

DISCRIMINANT=b%5E2-4ac

DISCRIMINANT=%28-1%29%5E2-4%281%29%280%29

DISCRIMINANT=1-0

DISCRIMINANT=1

Since the discriminant is a positive number,
there are exactly two real solutions.

Edwin

Answer by SAT Math Tutor(36) About Me  (Show Source):
You can put this solution on YOUR website!
Begin by setting up a quadratic equation equal to zero so:
35. 4m^2 - 20m + 25 = 0
45. x^2 - x = 0
Then, label a, b, and c which are the coefficients:
35. a = 4, b = -20, c = 25
45. a = 1, b = -1, c = 0
Then, use the discriminant formula b^2 - 4ac:
35. (-20)^2 - 4(4)(25) = 0
45. 1^2 - 4(1)(0) = 1