6a: 11x+9
6b: 5y+14
6c: -3c-10
6d: 15x+10y
6e: -2n+13n
6f: 5w-5z
6g: 19x+8y+9
6h: 11a+4b-5
hope this helps and is correct!!
Answer:
The pairs of integer having two real solution for
are
![a = -4, c = 5](https://tex.z-dn.net/?f=a%20%3D%20-4%2C%20c%20%3D%205)
![a = 1, c = 6](https://tex.z-dn.net/?f=a%20%3D%201%2C%20c%20%3D%206)
![a = 2, c = 3](https://tex.z-dn.net/?f=a%20%3D%202%2C%20c%20%3D%203)
![a = 3, c = 3](https://tex.z-dn.net/?f=a%20%3D%203%2C%20c%20%3D%203)
Step-by-step explanation:
Given
![ax^{2} -6x+c = 0](https://tex.z-dn.net/?f=ax%5E%7B2%7D%20-6x%2Bc%20%3D%200)
Now we will solve the equation by putting all the 6 pairs so we get the following
for ![a = -3 , c=-5](https://tex.z-dn.net/?f=a%20%3D%20-3%20%2C%20c%3D-5)
for ![a = -4 , c=5](https://tex.z-dn.net/?f=a%20%3D%20-4%20%2C%20c%3D5)
for ![a = 1 , c=6](https://tex.z-dn.net/?f=a%20%3D%201%20%2C%20c%3D6)
for ![a = 2 , c=3](https://tex.z-dn.net/?f=a%20%3D%202%20%2C%20c%3D3)
for ![a = 3 , c=3](https://tex.z-dn.net/?f=a%20%3D%203%20%2C%20c%3D3)
for ![a = 5 , c=4](https://tex.z-dn.net/?f=a%20%3D%205%20%2C%20c%3D4)
The above all are Quadratic equations inn general form ![ax^{2} +bx+c=0](https://tex.z-dn.net/?f=ax%5E%7B2%7D%20%2Bbx%2Bc%3D0)
where we have a,b and c constant values
So for a real Solution we must have
![Disciminant , b^{2} -4\timesa\timesc \geq 0](https://tex.z-dn.net/?f=Disciminant%20%2C%20b%5E%7B2%7D%20-4%5Ctimesa%5Ctimesc%20%5Cgeq%200)
for
we have
which is less than 0 ∴ not a real solution.
for
we have
which is greater than 0 ∴ a real solution.
for
we have
which is greater than 0 ∴ a real solution.
for
we have
which is greater than 0 ∴ a real solution.
for
we have
which is equal to 0 ∴ a real solution.
for
we have
which is less than 0 ∴ not a real solution.
Halting each time. 32, 16, 8
Please help!
Remember what we know about vertical angles and solve for x.
Answer: