Question

Can anyone please teach me how to solve for x in terms of a,b,c in this problem?
Thanks a lot!

ax^2+bx+c=0

Answer 1

Well, the general approach to solving equations of this kind is by using something known as a determinant.
The determinant of this equation is found by the following formula:

D=b²-4ac

(this formula is found by using Horners theorems on second degree equations, but I won't be going into depth on this subject)

There are 3 possible scenario's:

1)The determinant is negative (D<0): in this case there are no solutions for the problem.

2)The determinant is 0 (D=0): in this case there are two solutions, but both solutions have the same answer, so we only find 1 unique value as a solution

3)The determinant is positive (D>0): in this case there are two different solutions.

In cases 2 and 3, the two solutions can be calculated by using the equations:

x1 = (-b+sqrt(D))/(2a)
x2 = (-b-sqrt(D))/(2a)

We have now found the two possible solutions for the equation.

P.S. This is the most general approach for solving a second degree equation. Depending on the situation, there are possibly easier ways for solving, but you have to find that out for yourself.

Answer 2

Slove 4a=a+2d and -8=a+8d using substitution method