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