Root equation

In elementary algebra , root equations are determining equations in which the unknown (usually referred to as ) is at least once under a root . These can be square roots or roots with any root exponent. ${\displaystyle x}$

Many root equations can dissolve characterized in that a root isolated (alone leads to a page) and then the two sides of the equation with the root exponent (in the case of the square root So 2) potentiated . If necessary, repeat this process until all roots are eliminated.

It should be noted that raising an even number to the power is not an equivalence transformation . Such a calculation step can turn a false statement into a true statement, namely . Therefore, when raising to the power, false solutions can be added that are not solutions of the original equation. The test is therefore indispensable for root equations. ${\displaystyle 2=-2}$ ${\displaystyle 2^{2}=(-2)^{2}}$

example

Given is with equation

${\displaystyle {\sqrt {8-2x}}\,=\,1+{\sqrt {5-x}}}$

with the real numbers as the basic set . So all real numbers are searched that satisfy this equation. ${\displaystyle \mathbb {R} }$

First of all, the maximum permissible range of definition should be determined: The radicands of the two roots, i.e. the terms (arithmetic expressions) under these roots, must be positive or equal to 0. The condition is equivalent to . Accordingly is equivalent to . Both conditions must be met at the same time. We therefore get: ${\displaystyle D}$ ${\displaystyle 8-2x\geq 0}$ ${\displaystyle x\leq 4}$ ${\displaystyle 5-x\geq 0}$ ${\displaystyle x\leq 5}$

${\displaystyle D\,=\,\{x\in \mathbb {R} \mid x\leq 4\}={]{-\infty };4]}}$

To solve the equation, both sides of the equation are raised to the power (i.e. squared) with the root exponent (namely 2).

${\displaystyle \left({\sqrt {8-2x}}\right)^{2}\,=\,\left(1+{\sqrt {5-x}}\right)^{2}}$

On the right side is now the square of a sum . The binomial formula must therefore be used. ${\displaystyle (a+b)^{2}\,=\,a^{2}+2ab+b^{2}}$

${\displaystyle 8-2x\,=\,1^{2}+2\cdot 1\cdot {\sqrt {5-x}}+\left({\sqrt {5-x}}\right)^{2}}$

${\displaystyle 8-2x\,=\,1+2{\sqrt {5-x}}+(5-x)}$

Before you squar the equation on both sides again, you have to isolate the summand that contains the remaining root. This is done by subtracting and from both sides of the equation . ${\displaystyle 1}$ ${\displaystyle (5-x)}$

${\displaystyle 8-2x-1-(5-x)\,=\,2{\sqrt {5-x}}}$

${\displaystyle 8-2x-1-5+x\,=\,2{\sqrt {5-x}}}$

${\displaystyle 2-x\,=\,2{\sqrt {5-x}}}$

In order to remove the root that is still present, one squares both sides of the equation again.

${\displaystyle (2-x)^{2}\,=\,\left(2{\sqrt {5-x}}\right)^{2}}$

${\displaystyle 2^{2}-2\cdot 2\cdot x+x^{2}\,=\,4(5-x)}$

${\displaystyle 4-4x+x^{2}\,=\,20-4x}$

This equation is simplified by adding to ${\displaystyle 4x}$

${\displaystyle 4+x^{2}\,=\,20}$

and by subtracting the number 4 to

${\displaystyle x^{2}\,=\,16}$.

The last ( quadratic ) equation has two solutions:

${\displaystyle x_{1}=4;\quad x_{2}=-4}$

According to the remark made above, a sample is necessary. For results by substituting into the original equation: ${\displaystyle x=4}$

Left side: ${\displaystyle {\sqrt {8-2\cdot 4}}\,={\sqrt {8-8}}={\sqrt {0}}=0}$

Right side: ${\displaystyle 1+{\sqrt {5-4}}=1+{\sqrt {1}}=1+1=2}$

This turns out to be a sham solution. For , on the other hand, you get: ${\displaystyle x=4}$ ${\displaystyle x=-4}$

Left side: ${\displaystyle {\sqrt {8-2\cdot (-4)}}\,={\sqrt {8-(-8)}}={\sqrt {16}}=4}$

Right side: ${\displaystyle 1+{\sqrt {5-(-4)}}=1+{\sqrt {9}}=1+3=4}$

This solution therefore satisfies the original equation. The solution set of the equation is found: ${\displaystyle L}$

${\displaystyle L\,=\,\{-4\}}$

See also

• Solving equations

Web links

• Completely calculated examples (141 pages, PDF; 880 KB)