Quadratic Equations: Quadratic Equations with one Unknown
The notion of quadratic equation
Quadratic equation
A quadratic equation with unknown #x# is an equation of the form #ax^2+bx+c = 0#.
If #a=0# , then we are dealing with a linear equation, which we already discussed on how to solve it.
Note the subtle difference: in the linear case we also described the equation #0\cdot x+3=0# or #3=0#, as linear. Here we do not describe the equation #0x^2+2x+3=0# or #2x+3=0# quadratic.
In case you are familiar with the notion of function: the solutions of the equation #ax^2+bx+c=0# are zeros of the quadratic function #ax^2+bx+c#.
As with the linear case, with quadratic equations we also refer to equations that become quadratic after expanding the brackets and moving all terms to the left hand side. Consequently #2(x+1)^2=9x-4# and #3x^2=(x+1)\cdot(x-6)# are also called quadratic equations.
Mentioned below are a few useful rules for solving quadratic equations. We will deal with the famous abc-formula in the theory of General solution of a quadratic equation . We recall that two equations are equivalent if they have exactly the same solutions.
Consider the quadratic equation #ax^2+bx+c = 0# with unknown #x#.
- The equation #ax^2+bx+c = 0# is equivalent to #x^2+\frac{b}{a}x+\frac{c}{a}=0# .
- The solution of the equation #x^2+c=0# is \[ \begin{cases}x=\sqrt{-c}\lor x=-\sqrt{-c}&\text{if }c\lt0\\ x={0}&\text{if }c=0\\ none&\text{if }c\gt0\end {cases}\]
- If #x=p# is a solution of #ax^2+bx+c=0# , then #x=-p-\frac{b}{a}# is also a solution and \[ax^2+bx+c = a(x-p)\cdot(x-q)\tiny,\] is true, in which #q=-p-\frac{b}{a}#.
1. We can rewrite #ax^2+bx+c # as #a\cdot \left(x^2+\frac{b}{a}x+\frac{c}{a}\right)#. Because #a\ne0# follows from the zero divisors theorem that #x^2+\frac{b}{a}x+\frac{c}{a}=0# if and only if #ax^2+{b}x+{c}=0#. This proves the equivalence.
2. The equation #x^2+c=0# is equivalent to #x^2=-c#.
- If #c\gt0#, the right hand side is negative and the left hand side (a square) is non-negative, hence there are no solutions.
- If #c=0# , the zero divisors theorem states that the equation #x^2=0# can only have the solution #x=0#.
- Finally, if #c\lt0#, according to the theory Roots of an integer there is exactly one non-negative real number #p#, in such a way that #p^2 = -c#. This number #p# is written as #\sqrt{-c}# and is called the root of #-c#. Accordingly there is exactly one non-positive number #q# with#q^2=-c#, being #q=-\sqrt{-c}#. Therefore the solution of #x^2=-c# is #x=\sqrt{-c}\lor x= -\sqrt{-c}#.
3. When #x=p# is a solution of #ax^2+bx+c=0#, then by definition #ap^2+bp+c=0# is satisfied. We use this in the form #-ap^2-bp=c# to reduce the following: \[ax^2+bx+c=a\left(x-p\right)\cdot \left(x-q\right)\] where #q=-p-\frac{b}{a}#. After all, \[ \begin{array}{rcl}a\left(x-p\right)\cdot \left(x-q\right)&=&ax^2-apx-aqx+apq\\&&\phantom{xyzuvwxyzuvwxyz}\color{blue}{\text{expanding the brackets }}\\&=&ax^2-apx-a\left(-p-\frac{b}{a}\right)x+ap\left(-p-\frac{b}{a}\right)\\\\&&\phantom{xyzuvwxyzuvwxyz}\color{blue}{q=-p-\frac{b}{a}\text{ entered}}\\&=&ax^2-apx+apx+bx-ap^2-bp\\&&\phantom{xyzuvwxyzuvwxyz}\color{blue}{\text{expanding the brackets}}\\&=&ax^2+bx-ap^2-bp\\&&\phantom{xyzuvwxyzuvwxyz}\color{blue}{\text{simplified}}\\&=&ax^2+bx+c\\&&\phantom{xyzuvwxyzuvwxyz}\color{blue}{-ap^2-bp=c\text{ used}}\end {array}\]
#\begin{array}{rclcl}s^2+35&=&35\\&&\phantom{xxx}\blue{\text{the given equation}}\\ s^2&=&0\\&&\phantom{xxx}\blue{\text{constant terms to the right hand side}}\\ s&=&0\\&&\phantom{xxx}\blue{\text{according to the theory}}\end {array}#
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