No. A polynomial equation in one variablel ooks like P(x) = Q(x), where P and Q are polynomials.
Consider polynomial equations x^2 = 3 and x^2 = 1.
Obviously they have real solutions.
Subtract the two polynomial equations:
(x^2 - x^2) = (3 - 1)
0 = 2...
We get the polynomial equation 0 = 2. We call this a polynomial equation because single constants are also by definition polynomials.
Obviously 0 = 2 has no real solution.
Answer:
p-q= -4
p=q-4
x^2 + px + q=0
=> x^2 + (q-4)x + q = 0
d= b^2-4ac
=(q-4)^2 - 4(1)(q)
= q^2 + 16 -8q - 4q
=q^2 -12q+16
given that D=16
q^2 -12q+16 = 16
=> q^2 -12q = 0
=> q^2 = 12q
=> q = 12q/q
=> q= 12
Therefore the value of q is 12 and the value of p is q-4 = 12-4= 8
Hope you understood............
Answer:
The answer is WhAt ThE dOg DoInG
Step-by-step explanation:
The reason why it is this answer is because it is i am like that because i am smart
A=60
B=52
C=152
I’ve attached my work on how I got these answers
Hope this helps!
