Question

Can someone help me solve the following system of equations. Please explain how.

20a + b = -3
b^2 = 4ac
100a + 10b + c = 10

As you may see I'm trying to find a parabola.

Answer 1

b from the first equation b=-3-20a and substitute in second

(-3-20a)^2 =4ac
9 -2(-3)(-20a)+40a^2 =4ac
9+6(-20a)+40a^2 =4ac
9-120a+40a^2 =4ac

Answer 2

re-write 20a + b = -3 into the form b = -20a - 3 (1)

re-write 100a + 10b + c=10 as c = 10 - 100a - 10b (2)

Substitute 1 into 2 c = 10 - 100a -10(-20a - 3)

c = 100a - 20 (3)



substitute it into the equation (1) and (3) into b^2 = 4ac then (-20a -3)^2 = 4a(100a - 20)

expand, collect like terms and then solve for a

Answer 3

The answers are not integers.\[\left\{a=\frac{9}{40},b=-\frac{15}{2},c=\frac{125}{2}\right\} \]

Answer 4

b=-3-20a
Applying in the 2nd eqn
(-3-20a)^2=4ac
9+120a+400a^2=4ac
Applying in the 3rd eqn
100a-30-200a+c=10
c=100a+40
applying in 9+120a+400a^2=4ac
9+120a+400a^2=4a(100a+40)
9+120a+400a^2=400a^2+160a
9=40a
a=9/40
b=-3-9/2
=-15/2
c=125/2