Question

This question came up in my math test and I don't know how to solve it! Please help!

For the quadratic equation, x^2 - 4x + c = 0 where the equation has only one solution... Find the value of c.

Answer 1

\[ax^2+bx+c=0\]\[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]

\[x=\frac{4\pm\sqrt{16-4c}}{2}\]\[x=2\pm\sqrt{4-c}\]

Answer 2

Ok

Answer 3

there will be only one solution when \(x=2\pm0\)

Answer 4

Alright I get it thanks

Answer 5

A quadratic equation will have one solution when when and only when the discriminant equals zero. Therefore

\(b^2 - 4ac = 0\)

a = 1
b = -4
c = c

\((-4)^2 - 4(1)c = 0 \\ 16 - 4c = 0 \\ 16 = 4c \\ c = 4\)

Answer 6

@jadecat