Remember that quadratic formula:
![x= \frac{-b(+/-) \sqrt{b^{2}-4ac} }{2a}](https://tex.z-dn.net/?f=x%3D%20%5Cfrac%7B-b%28%2B%2F-%29%20%5Csqrt%7Bb%5E%7B2%7D-4ac%7D%20%7D%7B2a%7D%20)
is used to solve quadratic equations of the form
![ax^{2}+bx+c](https://tex.z-dn.net/?f=ax%5E%7B2%7D%2Bbx%2Bc)
.
Four our problem, the first one,
![(x-1)^2=4](https://tex.z-dn.net/?f=%28x-1%29%5E2%3D4)
, is not in the form
![ax^{2}+bx+c](https://tex.z-dn.net/?f=ax%5E%7B2%7D%2Bbx%2Bc)
. So is not a good idea to use the quadratic formula for this one.
The second one,
![0.25x^{2}+0.8x-8=0](https://tex.z-dn.net/?f=0.25x%5E%7B2%7D%2B0.8x-8%3D0)
, is in the correct form. Notice that in this equation
![a=0.25](https://tex.z-dn.net/?f=a%3D0.25)
,
![b=0.8](https://tex.z-dn.net/?f=b%3D0.8)
, and
![c=-8](https://tex.z-dn.net/?f=c%3D-8)
.
It would be appropriate to use the quadratic formula to solve this one.The third one,
![3x^{2}-4x=15](https://tex.z-dn.net/?f=3x%5E%7B2%7D-4x%3D15)
, can be expressed as:
![3x^{2}-4x-15=0](https://tex.z-dn.net/?f=3x%5E%7B2%7D-4x-15%3D0)
. This equation is also in the correct form. Here
![a=3](https://tex.z-dn.net/?f=a%3D3)
,
![b=-4](https://tex.z-dn.net/?f=b%3D-4)
, and
![c=-15](https://tex.z-dn.net/?f=c%3D-15)
.
It would be appropriate to use the quadratic formula to solve this one.The fourth one,
![(x-6)(x+6)=0](https://tex.z-dn.net/?f=%28x-6%29%28x%2B6%29%3D0)
, is not in the form
![ax^{2}+bx+c](https://tex.z-dn.net/?f=ax%5E%7B2%7D%2Bbx%2Bc)
. So is not a good idea to use the quadratic formula for this one.
We can conclude that you should use the quadratic formula to solve the
second and
third equations.
Answer:
Please see attachment
Step-by-step explanation:
Please see attachment
Answer:
the answer I believe will be 25.16
Y = x + 6
y has to be greater than or equal to 6
x has to be greater than or equal to 0
5000+850302=855302
Hope that helped:D