Math help...quadratic "tricky" functions?

Question

(2p-1)x^2 + (p+1)x + p-4 = 0

Find p such that the above equation has only one solution.

Thanks and please show me the steps.

Answer 1

(p+1)^2-4(p-4)(2p-1)=0

solve it ! do it.

and get p !

Answer 2

With more than one variable in an equation there has to be more than one answer.

Answer 3

a quadratic equation Ax^2 + Bx + C = 0 has only one solution if: B^2 - 4AC = 0

So, to solve for p:

(p+1)^2 - 4(2p-1)(p-4) = 0

or, p^2 + 2p + 1 - 4(2p^2 - 9p + 4) = 0

or, p^2 + 2p + 1 - 8p^2 + 36p - 16 = 0

or, -7p^2 + 38p - 15 = 0

or, 7p^2 - 38p + 15 = 0

solve for p to get the ans