Solve this quadratic equation b2 - 10b plus 25 equals 0
b^2 - 10b + 25 = 0
(b -5)(b - 5) = 0
Can you work it from there?
How to solve 2nd degree equations easily:
Ax^2+Bx+c=0
x=(-B+/-sqrt(B^2-4AC))/2A
I don't agree Anonymous.
To solve 2nd degrees equations (aka find the roots of a trinomial function), you have first to put it to this form :
ax² + bx + c = 0 (where a, b and c are reals)
Then you have to calculate its discriminant (its delta) :
Δ = b² - 4ac
If Δ < 0, there is no real solution.
If Δ = 0, there is one solution :
x = -b/(2a)
If Δ > 0, there are two solutions :
x = (-b + sqrt(Δ))/(2a)
or x = (-b - sqrt(Δ))/(2a)
To solve the quadratic equation b^2 - 10b + 25 = 0, we can use the quadratic formula. The quadratic formula is given by:
b = (-b ± √(b^2 - 4ac)) / (2a)
In this equation, a = 1, b = -10, and c = 25. Substituting these values into the formula, we get:
b = (-(-10) ± √((-10)^2 - 4(1)(25))) / (2(1))
Simplifying further:
b = (10 ± √(100 - 100)) / 2
b = (10 ± √0) / 2
Since √0 = 0, we have:
b = (10 ± 0) / 2
Therefore, b = 10/2 = 5.
Hence, the solution to the quadratic equation b^2 - 10b + 25 = 0 is b = 5.