Quadratic Equations

From wikiscience101
Jump to: navigation, search

In Algebra we learned the solution to the quadratic equation

ax^{2}+bx+c=0

is

x={\frac  {-b\pm {\sqrt  {b^{2}-4ac}}}{2a}}

But, if you are like me, we did not learn how to derive that solution. Read on to see one approach.

Simplify

To start, simplify the quadratic equation by dividing both sides by 'a'. This does not change the values of x that solve this equation. The equeation becomes

x^{2}+(b/a)x+c/a=0

Replace b/a with B and c/a with C, so now we are looking for the solution to the following.

x^{2}+Bx+C=0

Factoring

This simplified quadratic equation

x^{2}+Bx+C=0

can be factored to

(x-k_{1})*(x-k_{2})=0

where k_{1} and k_{2} are the two solution values for x. Multiplying this out, we have

x^{2}+(-k_{1}-k_{2})x+k_{1}k_{2}=0

So, to derive the solution to the simplified quadratic equation, we need to solve the following simultaneous equations for k_{1} and k_{2} as a function of B and C.

B=-k_{1}-k_{2}
C=k_{1}*k_{2}

First Try

This looks easy. Lets solve for k_{1} using the first equation, and substitute the solution for k_{1} into the second equation.

C=(-B-k_{2})*k_{2}

simplifying, we have

k_{2}^{2}+Bk_{2}+C=0

Oh no, we are back to solving the original problem. This approach is a dead end.

Second Try

Remember, we are trying to solve the following equations for k_{1} and k_{2}:

B=-k_{1}-k_{2}
C=k_{1}*k_{2}

Let's create a new equation by squaring both sides of the first equation and then substituting C for k_{1}*k_{2}.

B^{2}=k_{1}^{2}+k_{2}^{2}+2k_{1}k_{2}
B^{2}-2C=k_{1}^{2}+k_{2}^{2}

And, now let's solve these 2 equations using geometry.

B^{2}-2C=k_{1}^{2}+k_{2}^{2}
k_{2}=-k_{1}-B

Referring to the following diagram, the blue triangle represents the first equation, and the green line represents the second.

Quadratic.svg

Regarding this daigram, we know what P and Q are. This leads to R, which leads to S, which leads finally to k_{2}.

P^{2}=k_{1}^{2}+k_{2}^{2}=B^{2}-2C
Q^{2}=B^{2}/2
R={\sqrt  {P^{2}-Q^{2}}}={\sqrt  {B^{2}-2C-B^{2}/2}}={\sqrt  {B^{2}/2-2C}}
S={\sqrt  {B^{2}/2-2C}}/{\sqrt  {2}}
k_{2}=-B/2-{\sqrt  {B^{2}/2-2C}}/{\sqrt  {2}}

This can be simplified to

k_{2}={\frac  {-B-{\sqrt  {B^{2}-4C}}}{2}}

k_{1} can be determined from k_{2}

k_{1}=-B-k_{2}={\frac  {-2B}{2}}-{\frac  {-B-{\sqrt  {B^{2}-4C}}}{2}}
k_{1}={\frac  {-B+{\sqrt  {B^{2}-4C}}}{2}}

We now have the two values of x that solve the simplified quadratic equation.

x^{2}+Bx+C=0
x={\frac  {-B\pm {\sqrt  {B^{2}-4C}}}{2}}

Replacing B with b/a, and C with c/a we have the solution to the quadratic equation

ax^{2}+bx+c=0
x={\frac  {-b\pm {\sqrt  {b^{2}-4ac}}}{2a}}