How do I do these discriminant maths question?

For a quadratic equation of the form ax² + bx + c = 0, you look at the value(s) of √(b² - 4ac), or even simpler, look at the value of b² - 4ac. If this value is greater than zero, the equation will have real and distinct roots. If this value is equal to zero, there will be one real root (although is is called "doubled.") If this value is less than zero, there will be no real roots.

For #1, we want to choose values for k such that k² - 64 is ≥ 0. Thus k² ≥ 64, k ≥ 8 or k ≤ -8

For #2, we want to choose values for m such that 4m² - 36 < 0. Thus 4m² < 36, m² < 9, -3 < m < 3.

For #3, we want to choose values for p such that 4 - 4p² > 0. Thus 4 > 4p², p² < 1, -1 < p < 1 (but not zero).

ax²+bx+c=0 --> discriminant = b² - 4ac

1/

discriminant = k² - 4(16)= k² - 64

the equation x²+kx+16=0 has real roots ---> k² - 64 ≥ 0 , find k

2/

the equation x²+2mx+9=0 has no real roots --->discriminant = 4m^2-36 < 0 , find m

3/

the equation px²+2x+p=0 has real and distinct roots ---> discriminant =4-p^2 > 0 , find p

1. d ≥ 0

k^2 - 4(1)(16) ≥ 0

k^2 - 64 ≥ 0

(k - 8)(k + 8) ≥ 0

k ≥ 8 or k ≤ -8

2. d < 0

(2m)^2 - 4(1)(9) < 0

(2m)^2 - 36 < 0

(2m - 6)(2m + 6) < 0

(m - 3)(m + 3) < 0

-3 < m < 3

3. d > 0

2^2 - 4(p)(p) > 0

4 - 4p^2 > 0

4(1 - p)(1 + p) > 0

p > 1 or p < -1