Respuesta :

[tex]\bf 5-\sqrt{x+10}=\sqrt{7-x}\impliedby \textit{squaring both sides} \\\\\\ (5-\sqrt{x+10})^2=(\sqrt{7-x})^2 \\\\\\ (5^2)-10\sqrt{x+10}+(x+10)=7-x \\\\\\ 35-10\sqrt{x+10}+x=7-x\implies 28-10\sqrt{x+10}+2x=0 \\\\\\ 2x+28=10\sqrt{x+10}\implies x+14=5\sqrt{x+10}\impliedby \begin{array}{llll} squaring\ both\\ sides\ again \end{array} \\\\\\ [/tex]

[tex]\bf (x^2)+28x+(14^2)=5^2(x+10)\implies x^2+28x+196=25x+250 \\\\\\ \begin{array}{lcclll} x^2&+3x&-54&=0\\ &\uparrow &\uparrow \\ &-6+9&-6\cdot 9 \end{array} \implies (x+9)(x-6)=0\implies \begin{cases} x+9=0\\ \boxed{x=-9}\\ -----\\ x-6=0\\ \boxed{x=6} \end{cases}[/tex]

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