Løsning af ligninger med kvadratrødder

$\begin{array}{rl}\sqrt{2x-7}-3& =4\\ \\ \sqrt{2x-7}& =7\\ \end{array}$‍

$\begin{array}{rl}\sqrt{2x-7}& =7\\ \\ {\left(\sqrt{2x-7}\right)}^2& =(7{)}^2\\ \\ 2x-7& =49\\ \\ 2x& =56\\ \\ x& =28\end{array}$‍

$\begin{array}{rl}\sqrt{2x-7}-3& =4\\ \\ \sqrt{2(28)-7}-3& \stackrel{?}{=}4\\ \\ \sqrt{49}-3& \stackrel{?}{=}4\\ \\ 7-3& \stackrel{?}{=}4\\ \\ 4& =4\end{array}$‍

$\begin{array}{rl}6+3x& =\sqrt{2x+12}+2x\\ \\ 6+x& =\sqrt{2x+12}\\ \end{array}$‍

$\begin{array}{rl}6+x& =\sqrt{2x+12}\\ \\ {(6+x)}^2& ={\left(\sqrt{2x+12}\right)}^2\\ \\ 36+12x+x^2& =2x+12\\ \\ x^2+10x+24& =0\\ \\ (x+6)(x+4)& =0\end{array}$‍

$\begin{array}{rl}6+3x& =\sqrt{2x+12}+2x\\ \\ 6+3(-6)& \stackrel{?}{=}\sqrt{2(-6)+12}+2(-6)\\ \\ -12& \stackrel{?}{=}\sqrt{0}-12\\ \\ -12& \stackrel{?}{=}0-12\\ \\ -12& =-12\end{array}$‍

$\begin{array}{rl}6+3x& =\sqrt{2x+12}+2x\\ \\ 6+3(-4)& \stackrel{?}{=}\sqrt{2(-4)+12}+2(-4)\\ \\ -6& \stackrel{?}{=}\sqrt{4}-8\\ \\ -6& \stackrel{?}{=}2-8\\ \\ -6& =-6\end{array}$‍

$\begin{array}{rl}\sqrt{7x^2+15x}-2x& =5+x\\ \\ \sqrt{7x^2+15x}& =5+3x\\ \end{array}$‍

$\begin{array}{rl}\sqrt{7x^2+15x}& =5+3x\\ \\ {\left(\sqrt{7x^2+15x}\right)}^2& ={(5+3x)}^2\\ \\ 7x^2+15x& =25+30x+9x^2\\ \\ 0& =2x^2+15x+25\\ \\ 0& =(2x+5)(x+5)\end{array}$‍

$\begin{array}{rl}\sqrt{7x^2+15x}-2x& =5+x\\ \\ \sqrt{7{\left(-5\right)}^2+15\left(-5\right)}-2\left(-5\right)& \stackrel{?}{=}5+\left(-5\right)\\ \\ \sqrt{100}+10& \stackrel{?}{=}0\\ \\ 10+10& \stackrel{?}{=}0\\ \\ 20& \ne 0\\ \end{array}$‍

$\begin{array}{rl}\sqrt{7x^2+15x}-2x& =5+x\\ \\ \sqrt{7{\left(-{\displaystyle \frac{5}{2}}\right)}^2+15\left(-{\displaystyle \frac{5}{2}}\right)}-2\left(-{\displaystyle \frac{5}{2}}\right)& \stackrel{?}{=}5+\left(-{\displaystyle \frac{5}{2}}\right)\\ \\ \sqrt{{\displaystyle \frac{175}{4}}-{\displaystyle \frac{75}{2}}}+5& \stackrel{?}{=}{\displaystyle \frac{5}{2}}\\ \\ \sqrt{{\displaystyle \frac{25}{4}}}+5& \stackrel{?}{=}{\displaystyle \frac{5}{2}}\\ \\ {\displaystyle \frac{5}{2}}+5& \stackrel{?}{=}{\displaystyle \frac{5}{2}}\\ \\ {\displaystyle \frac{15}{2}}& \ne {\displaystyle \frac{5}{2}}\\ \end{array}$‍