I did not get any zeros since the graph doesn’t cross the x axis, meaning that there are no rational zeros
However, here is the method u can use to find the zeros lol
You can use the quadratic formula in order to get the zeros
This is the equation therefore use these values
ax^2+bx+c=0
A=1
B= -5
C=12
The quadratic formula is -b±√(b^2-4ac))/2a (I left a picture just in case)
Answer:
A. 2 distinct roots.
Step-by-step explanation:
2x^2 + 8x + 3 = 0
Finding the discriminant:
b^2 - 4ac = 8^2 - 4 * 2 * 3
= 64 - 24
= 40
The discriminant is positive but not a perfect square
So there are 2 distinct ,real, irrational roots.
Simplifying
2c + 3 = 3c + -4
Reorder the terms:
3 + 2c = 3c + -4
Reorder the terms:
3 + 2c = -4 + 3c
Solving
3 + 2c = -4 + 3c
Solving for variable 'c'.
Move all terms containing c to the left, all other terms to the right.
Add '-3c' to each side of the equation.
3 + 2c + -3c = -4 + 3c + -3c
Combine like terms: 2c + -3c = -1c
3 + -1c = -4 + 3c + -3c
Combine like terms: 3c + -3c = 0
3 + -1c = -4 + 0
3 + -1c = -4
Add '-3' to each side of the equation.
3 + -3 + -1c = -4 + -3
Combine like terms: 3 + -3 = 0
0 + -1c = -4 + -3
-1c = -4 + -3
Combine like terms: -4 + -3 = -7
-1c = -7
Divide each side by '-1'.
c = 7
Simplifying
c = 7
Hi the answer to your question is p=r-9/5 i hope this helps.
You would need to substitute a number for y. I would probably put 0 for y which would then give you x.
y = 0
x = 1
Then you can easily graph the point. :)
