Quadratic Equations

Equation of form `ax^2+bx+c=0` where a,b,c are real numbers and `a!=0` (if a=0 then equation becomes linear: `bx+c=0`), is called quadratic. If a=1, then quadratic equation is called reduced, if `a!=1` - equation is called unreduced. a is called first coefficient, b is called second coefficient, c is free member.

Roots of the quadratic equation can be found using formula `x=(-b+-sqrt(b^2-4ac))/(2a)`.

Expression `D=b^2-4ac` is called discriminant of the quadratic equation.

There are three possible cases:

  1. If D<0, equation doesn't have roots;
  2. If D=0, equation has one root; it is also said that equation has two equal roots;
  3. If D>0, equation has two roots.

Now, formula for roots of the equation can be rewritten as `x=(-b+-sqrt(D))/(2a)`.

If b=2k, then `x=(-k+-sqrt(k^2-ac))/a` , where `k=b/2`. This formula is especially useful when b is even, i.e. k is integer.

Example 1. Solve equation `2x^2-5x+2=0`.

Here `a=2,b=-5,c=2`. Now find discriminant: `D=b^2-4ac=(-5)^2-4*2*2=9`.

Since D>0 then equation has two roots: `x=(-(-5)+-sqrt(9))/(2*2)=(5+-3)/4`.

So, `x_1=(5+3)/4=2` and `x_2=(5-3)/4=1/2` are roots of given equation.

Example 2. Solve `x^2-6x+9=0`.

Here `a=1,b=-6,c=9`. Since b is even we use formula for roots with k: `k=b/2=(-6)/2=-3`.

`x=(-k+-sqrt(k^2-ac))/a=(-(-3)+-sqrt((-3)^2-1*9))/1=3+-sqrt(0)=3`.

Thus, the only root of the equation is `x=3`.

Example 3. Solve `2x^2-3x+5=0`.

Here `a=2,b=-3,c=5`. Now find discriminant: `D=b^2-4ac=(-3)^2-2*5=-1`.

Since D<0, equation doesn't have roots.