Linear Equations with a Single Unknown: Variations
Solving by factorization
Zero divisors
If the product of two real numbers is zero, then at least one of both numbers is equal to zero. The other way around is of course also true that the product of two numbers is zero if at least one of both is zero.
In other words: for each pair of real numbers #a# and #b# holds #a\cdot b = 0# then and only then if #a=0# of #b=0#.
Because #0\cdot a = 0# is a known law, we only have to show that if #a\cdot b = 0#, it is also true that #a=0# or #b=0#.
Assume that #a\ne0#. Then #\frac{1}{a}# is also a real number. This number satisfies #\frac{1}{a}\cdot a = 1#, hence #b = \frac{1}{a} \cdot a \cdot b = \frac{1}{a}\cdot 0 = 0#. This shows that #b=0#. We conclude that #a=0# or #b=0#, from which the statement follows.
This fact has the following consequence for equations.
Solving by factorization
Let #a#, #b#, #c#, and #d# be real numbers.
- The solution of the equation #(x-a)\cdot (x-b) = 0# with unknown #x# is #x=a\lor x=b#.
- The equation #(ax+b)\cdot (cx+d) = 0# is equivalent to #ax+b=0\lor cx+d=0#.
The first statement is a special case of the second. The second statement is a direct consequence of the above statement Zero divisors.
Expansion of the brackets in #(x-a)\cdot (x-b) = 0# leads to the equation #x^2-(a+b)\cdot x+a\cdot b=0#. This is a quadratic equation, for which a solution method will be discussed later. Thus, an equation of the form #x^2-p\cdot x+q=0# has solution #x=a\lor x=b# if #a+b=p# and #a\cdot b = q#.
In special cases we can find these #a# and #b#; for instance if they are integers. For example, the equation #x^2+4x-12=0#, the case with #p=-4# and #q=-12#, can be solved by looking for integers #a# and #b# with #a+b=-4# and #a\cdot b=-12#. By checking all divisors of #12#, we find that #a=-6# and #b=\frac{-12}{-6}=2# satisfy the equations #a+b= -4# and #a\cdot b = -12#, so the original equation is equivalent to #(x+6)\cdot(x-2)=0#, and hence to #x=-6\lor x= 2#.
This means we can also handle equations like #(3x+4)\cdot(5x-9) = 0#.
For, #(3 x -5)\cdot ( 10 x +3) = 0# is equivalent to #3 x -5=0\lor 10 x +3 = 0#. The first of these two linear equations with unknown #x# has solution #x={{5}\over{3}}# and the second #x=-{{3}\over{10}}#. Hence the answer is #x= {{5}\over{3}}\lor x = -{{3}\over{10}}#.
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