The equation
(p-1)x^2 + 4x + (p-5)=0, where p is a constant,
has no real roots.
(a) Show that p satisfies p^2 - 6p+1>0.

Respuesta :

Answer:

Proof is shown below.

Step-by-step explanation:

Use quadratic Formula, where from the standard form ax^2 + bx + c = 0, where a is not zero, we identify a = p-1, b=4, c = p-5, and since the given equation has no real roots, the discriminant b^2 - 4ac < 0, so that substituting we have that,

4^2 - 4(p-1)(p-5) < 0

16 - 4(p^2 - 6p + 5) < 0

16 - 4p^2 +24p - 20 < 0

-4p^2 + 24p - 4 < 0,

p^2 - 6p + 1 > 0, by dividing both sides of the inequality by -4.