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preety5445
22.07.2021 •
Mathematics
Which of the following could be appropriate values for an equation to be in Standard Form (Ax + By = C)? A=14,B=34 and C=0 A=2,B=12 and C=0 A=−2,B=3 and C=5 A=1,B=0 and C=5
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Ответ:
(0,0)
(1/15,1/15)
Please let me know if you are expected to find complex solutions rather than just the real solutions I have provided.
Thanks kindly.
Step-by-step explanation:
Your derivatives look good.
Now we have to solve this system:
45x^2-3y=0
-3x+45y^2=0
Solving the first equation for y in terms of x:
45x^2-3y=0
Add 3y on both sides:
45x^2=3y
Divide both sides by 3:
15x^2=y
We are going to plug this into second equation:
-3x+45y^2=0 with y=15x^2
-3x+45(15x^2)^2=0
-3x+45×225x^4=0
Divide both sides by -3:
x-15×225x^4=0
x-3375x^4=0
Factor left hand side:
x(1-3375x^3)=0
This implies that x=0 or 1-3375x^3=0.
We can factor left hand side of second equation. It is a difference of cubes.
(1-15x)(1+15x+225x^2)=0
This means x=1/15 or 1+15x+225x^2=0.
Let's solve that second equation just a line above this.
1+15x+225x^2=0
a=225
b=15
c=1
First, I will find the discriminant to find out if the solution is even real.
b^2-4ac
(15)^2-4(225)(1)
225-4(225)
225(1-4)
225(-3)
A positive times a negative is negative so the discriminant tells us even if we find the solution, it is not real.
So we had x=0, 1/15 that we need to look at. I'm going to use the earlier equation I solved for y in terms of x to find the corresponding values per each x-solution.
15x^2=y with x=0:
15(0)^2=y
15(0)=y
0=y
So we have point (0,0) is a solution to the system.
15x^2=y with x=1/15:
15(1/15)^2=y
15/15^2=y
1/15=y
So we also have the point (1/15,1/15) is a solution to the system.