Question

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

Answer 1

Answer:

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

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x = 0

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x - 1 = 0

x - 1 + 1 = 0 + 1    

x = 1

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Answer 2

Answer:

So, the values of z for which the equation has only one real solution are        5, 10

Step-by-step explanation:

If z=5, then the equation becomes the linear equation -5x+5=0, which has one real solution, x=1.

For any other value of z, the equation is quadratic. For a quadratic to have only one real solution, the discriminant must be 0. Therefore, we have