A kosarad üres
Solving equations with examining the domain and the range
Examining the domain of the equation
We often come across equations for which the fundamental set – in which we look for the solutions – is not given. In this case we have to give the largest set on which the equation has a meaning.
The conditions resulting this way can often make it easier to find the solutions.
Example 1
Let us solve the following equation.
\latex{\sqrt{x-2} =\sqrt{2-x}}
Solution
Since under the square root sign there can only be non-negative numbers, therefore the domain can contain only such real numbers for which \latex{x-2\geq 0} and \latex{2-x\geq 0} are fulfilled at once. And this is true only in the case of one real number: \latex{ x = 2 }.
We can check that this is also the solution of the equation.
Example 2
Let us find the value of parameter \latex{a} so that the domains of the following equations
will not be the empty set.
will not be the empty set.
- \latex{\sqrt{x+2}=\sqrt{-x+a}}
- \latex{\sqrt{x+2}=\sqrt{x+a}}
Solution (a)
The conditions because of the square root sign are as follows: \latex{x+2\geq 0} and
\latex{-x+a\geq 0}. It implies that \latex{x\geq -2} and \latex{x\leq a}, i.e. \latex{-2\leq x\leq a}. This interval is not
empty only if \latex{a\geq -2}. (Figure 6)
\latex{-x+a\geq 0}. It implies that \latex{x\geq -2} and \latex{x\leq a}, i.e. \latex{-2\leq x\leq a}. This interval is not
empty only if \latex{a\geq -2}. (Figure 6)
Figure 6
\latex{ -2 }
\latex{ a }
Solution (b)
Similarly to case a): \latex{x+2\geq 0} and \latex{x+a\geq 0}. When expressing \latex{x-2\leq x} and \latex{-a\leq x}. Depending on the value of parameter a one of the intervals surely contains the other one. Thus for any real value of parameter a the domain is not empty.
The domain (Figure 7):
when \latex{ a = } 2 then \latex{-2\leq x};
when \latex{ a \lt 2 } then \latex{-a\leq x};
when \latex{ a\gt 2 } then \latex{-2\leq x}.
when \latex{ a \lt 2 } then \latex{-a\leq x};
when \latex{ a\gt 2 } then \latex{-2\leq x}.
Figure 7
If \latex{ a\gt 2 }
If \latex{ a\lt 2 }
\latex{ -2 }
\latex{ -a }
\latex{ -2 }
\latex{ -a }
Naturally based on this we cannot say yet that the equation will surely have a solution too. In the case of \latex{ a = 2 } we get an identity, which means that every real number of the domain will be a solution \latex{\left(-2\leq x\right) }. But if \latex{a\neq 2}, then we do not get a solution.
Examining the range
While solving equations it is often useful if we examine the range of the expressions on the two sides of the equality.
Example 3
Let us solve the following equation if \latex{x} and \latex{y} are real numbers.
\latex{\left|x+2\right| +\left|x-y\right| =0}
Solution
The sum of the two absolute values on the left side of the equation is obviously non-negative. It can be said about both of the expressions separately:
\latex{\left|x+2\right| \geq 0; \left|x-y\right| \geq 0}
This implies that the equality can be satisfied if and only if the values of both terms are zero.
It means that
It means that
\latex{x+2=0};
\latex{x-y=0}.
\latex{x-y=0}.
When solving the equations we get \latex{ x = -2 } and \latex{ y = -2 }. By checking we can confirm that this is indeed solution of the equation.
Example 4
Let us solve equation \latex{ (x - y)^2 + x^2 - 2x = -1 } on the set of real numbers.
Solution
Let us transform the equation as follows:
\latex{ (x-y)^2 + x^2-2x + 1 = 0. }
The last three terms of the left side make a complete square:
(x – y)2 + (x – 1)2 = 0
(x – y)2 + (x – 1)2 = 0
Since \latex{\left(x-y\right)^{2} \geq 0} and \latex{\left(x-1\right)^{2} \geq 0}, the sum of these will be equal to zero only if both expressions are equal to zero:
\latex{(x-y)^2 = 0};
\latex{(x-1)^2 = 0}.
\latex{x = y};
\latex{x = 1}.
\latex{(x-1)^2 = 0}.
\latex{x = y};
\latex{x = 1}.
Thus the solution is \latex{ x = 1 } and \latex{ y = 1 }.
Exercises
{{exercise_number}}. Solve the following equations.
- \latex{\sqrt{x-3}=\sqrt{2-x} }
- \latex{\sqrt{-2x-4}=\sqrt{-9+3x} }
- \latex{\sqrt{x-1}+\frac{1}{x-1} =\sqrt{1-x}}
- \latex{\frac{\sqrt{6-2x} }{x-3} =\sqrt{4x -12}}
{{exercise_number}}. For which real numbers a will the domain of the below equations be the empty set?
- \latex{\sqrt{x-7}=\sqrt{a-x} }
- \latex{\sqrt{2x-6}=\sqrt{a-x} }
- \latex{\sqrt{2x+4}=\sqrt{a-3x} }
- \latex{\sqrt{x}=\sqrt{2a-4x} }
{{exercise_number}}. Solve the following equations.
- \latex{\left|2x+1\right|+\left|x-2y\right| =0 }
- \latex{\left|3x-2y\right| +\sqrt{y-2} =0 }
- \latex{\sqrt{2x+3y} +\left(x+2\right)^{2} =0 }
- \latex{x^{2}-4x+\sqrt{2x-5y} =-4 }
- \latex{\sqrt{x-2}=2-x }
- \latex{\sqrt{3x-6}+\left|x+y\right|=-z^{2} +2z-1}
Puzzle
Numbers \latex{1-x;\; 2-x;\; 3-x;\dots;\; 100-x} are given.
Calculate their product if \latex{x = 77}. What will be their sum if \latex{x = 50.5?}
Calculate their product if \latex{x = 77}. What will be their sum if \latex{x = 50.5?}
