Question

Find the number of real number solutions for the equation. x2 – 10x + 25 = 0

2

1

those are my f answqer chgoices

Answer 1

There is one solution

Explanation:

x2 – 10x + 25 = 0

We can factor

(x-5) (x-5) = 0

Using the zero product property

x-5=0 x-5=0

x=5 x=5

There is one solution

Answer 2

Option B is correct

The number of real number solutions for the given equation is, 1.

Explanation:

Given the equation:
`x^2-10x+25 =0`

Since, this is a quadratic equation of the form of
`ax^2+bx+c =0`

where a =1 , b = -10 and c =25.

The discriminant of a quadratic equation is,
`{b^2-4ac}`

then;

Discriminant =
`{(-10)^2-4(1)(25)}` = (100-100) = 0

Since, a discriminant of zero means there is only one real solution for x.

`x = (-b\pm√(b^2-4ac))/(2a)`

Substitute the given values.

`x = (-(-10) \pm 0)/(2(1))`

`x = (10)/(2)`

Simplify:

x = 5

Therefore, the number of real solution for the given equation is, 1