Question

Determine the number of real solutions -4x^2+20x-25=0'

Answer 1

factor out the -1
-1(4x²-20x+25)=0
use ac method

4 times 25=100
what 2 factors of 100 add to -20?
-10 and -10

-1(4x²-10x-10x+25)=0
group
-1((4x²-10x)+(-10x+25))=0
factor
-1(2x(2x-5)+(-5)(2x-5))=0
reverse distribute, remember that ab+ac=a(b+c)
-1(2x-5)(2x-5)=0
set equal to 0
2x-5=0
2x=5
x=5/2

there is 1 real solution which is 5/2

Answer 2

To find out the number of real solutions, we use something called "the discriminant". This is usually given the symbol "Δ", pronounced "delta"

Δ = b^2 - 4ac, where a is the x^2 coefficient, b is the x coefficient and c is the integer

When Δ = 0, there is one real solution, when Δ < 0, there are no real solutions, and when Δ > 0, there are 2 real solutions

So, substituting the values in, we get:

Δ = (20)^2 - 4(-4)(-25) = 0

Therefore, there is one real solution to the polynomial given.

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Hope I helped!! xx