Find all real numbers z for which the equation (z-5)x^2-zx+5=0 only has one real solution. Help, Please?

1 Answer
Jan 25, 2018

=>z=10+-4sqrt(5)

Explanation:

In order for a quadratic equation in the form ax^2+bx+c=0 to have 1 solution, b^2-4ac=0

In this case:

  • a=z-5
  • b=-z
  • c=5

plug in the values to get:

(-z)^2-4(z-5)(5)=0

=>(-z)^2-4(5z-5)=0

=>z^2-20z+20=0

=>z=(-(-20)+-sqrt((-20)^2-4(1)(20)))/(2*1)

=>z=(20+-sqrt(400-80))/2

=>z=(20+-sqrt(320))/2

=>z=(20+-8sqrt(10))/2

=>z=10+-4sqrt(5)