Find all real numbers z for which the equation (z-5)x^2-zx+5=0 only has one real solution. i am giving

z = 5

x = 0

x = 1

Step-by-step explanation:

(z - 5)x^2 - (z - 5)x = 0          Take out z - 5 as a common factor

(z - 5)(x^2 - x)                        Take out x as a common factor

(z - 5)(x)(x - 1)                          Any one of these could equal 0. There are 3 solutons.

z - 5 = 0                                  Add 5 to both sides

z - 5 + 5 = 0 + 5

z = 5

x = 0

x - 1 = 0

x - 1 + 1 = 0 + 1    

x = 1

step-by-step explanation:

equation

you are asked to write the equation

  expression = 0

using the given expression. when you do that, you have

  2x^2-2x-12 = 0

factoring

you can see that each of the coefficients is even, so a factor of 2 can be removed immediately.

  2(x^2 -x -6) = 0

if you multiply the binomials (x+a)(x+b), you get

  (x +a)(x +b) = x^2 +(a+b)x +ab

this tells you the coefficient of x is the sum of two factors of the constant term. here, that means -1 is the sum of two factors of -6. we want to know those factors so we can factor the trinomial.

  -6 = -1×6 = -2×3 = -3×2 = -6×1

the factor pair that sums to -1 is {-3, 2}, so our factorization is

  2(x -3)(x +2) = 0

solutions

the solutions to this equation are the values of x that make the product zero. a product is zero only when only of the factors of the product is zero. here, that means

  x - 3 = 0

  x = 3 . . . add 3 to both sides of the equation

or

  x +2 = 0

  x = -2 . . . add -2 to both sides of the equation

the solutions to the equation are x=3 or x=-2. both of these values of x satisfy the equation—they make it be true.

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step-by-step explanation: