12. Quadratic Equations
Equations of the form $ax^2+bx+c=0$ are called quadratic equations.
Note how the degree of the highest term is 2 in these equations.
The easiest way to solve an equation of this form is to factorize the LHS into a product of two linear terms and solve from there.
e.g. solve: $x^2+5x+6=0$
We note $x^2+5x+6=(x+3)(x+2)=0$, so $(x+3)=0$ giving $x=-3$ or $(x+2)=0$ giving $x=-2$ for a solution set of {-2,-3}.
Since the equation has degree two, it has two roots, though they might not always be distinct.

Where it is not possible to factorize the quadratic as above, we use the quadratic formula to solve the equation.
This is as below:
If $ax^2+bx+c=0$, then $x={-b±√{b^2-4ac}}/{2a}$
e.g. solve: $x^2+5x+6=0$
Here, we note comparing with $ax^2+bx+c=0$ that $a=1,b=5,c=6$
So $x={-5±√{5^2-4*1*6}}/{2*1}={-5±√1}/2={-5±1}/2$
Giving $x=-2$ or $x=-3$ which is the same as the answer we obtained earlier.

So given a quadratic equation of the form $ax^2+bx+c=0$, what can we say about the roots?
1. the term under the square root sign of the quadratic formula is called the discriminant
   if the discriminant is > 0, the roots are real numbers.
   if the discriminant is < 0, the roots are complex numbers.
   if the discriminant is = 0, the roots are real and repeating, and are both equal to $-b/{2a}$

2. the sum of the roots of a quadratic equation = $-b/a$

3. the product of the roots of a quadratic equation = $c/a$

4. if the quadratic has integer coefficients and one root is $x={-b+√{b^2-4ac}}/{2a}$,
   then the other root has to be: $x={-b-√{b^2-4ac}}/{2a}$, and vice-versa.

e.g. Given a quadratic $2x^2+3x+1=0$, what can you say about the roots?
Here, $a=2,b=3,c=1$, roots are $({-3±√{3^2-4*2*1}})/{2*2}={-3±1}/4$
So $x=-1$ or $x=-1/2$, product of the roots=$1/2$=$c/a$, sum of the roots=$-3/2$=$-b/a$
Also, the discriminant is $({3^2-4*2*1})$=1 > 0, so the roots are real numbers.

 


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