Respuesta :
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.
Don't hesitate to ask further questions!
Hope I helped!! xx
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.
Don't hesitate to ask further questions!
Hope I helped!! xx
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
-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