In quadratic equation:
a x ^ 2 + b x + c
value D = b ^ 2 - 4ac is called the discriminant.
When Discriminant is zero we get just one real solution.
p x ^ 2 - 12 x + 4 = 0
In this case :
a = p , b = - 12 , c = 4
D = b ^ 2 - 4 a c =
( - 12 ) ^ 2 - 4 * p * 4 =
144 - 16 p = 0 Add 16 p to both sides
144 - 16 p + 16 p = 0 + 16 p
144 = 16 p Divide both sides by 16
144 / 16 = p
9 = p
p = 9
p x ^ 2 - 12 x + 4 = 0
become:
9 x ^ 2 - 12 x + 4 = 0
By the way the solution is x = 2 / 3
Find p such that px^2-12x+4=0 has only one solution.
Thank you!!!
2 answers
the answer 9
1 answer