Question

AYUDENMEEE
Pliss necesito resolver esta ecuación junto a su comprobación :(
$\sqrt{\sqrt{x+5}+x } =5$

Answer 1

$\sqrt{\sqrt{x+5}+x}=5\ (Eliminamos\ las\ raices) \\ \\ \sqrt{\sqrt{x+5}+x}^{2} =5^{2} \\ \\ \sqrt{x+5}^{2} +x}^{2} =25^{2} \\ \\ x+5 = 625 - x^{2} \\ \\ x = 20\\ \\ Comprobacion\\ \sqrt{\sqrt{20+5}+20}=5 \\ \\ \sqrt{\sqrt{25}+20}=5\\ \\ \sqrt{5+20}=5\\ \\ \sqrt{25}=5\\ \\ 5=5$

$\\\ Lo\ que\ realice\ en\ la\ fila\ 4\\ x+5 = 625 - x^{2} \\ x+x^{2}-620 = 0\\ \\ Formula\:general\\ x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ \\ Para\ :a=1,\:b=-51,\:c=620\\ x_{1,\:2}=\frac{-\left(-51\right)\pm \sqrt{\left(-51\right)^2-4\cdot \:1\cdot \:620}}{2\cdot \:1}\\ x_{1,\:2}=\frac{-\left(-51\right)\pm \:11}{2\cdot \:1}\\ x_1=\frac{-\left(-51\right)+11}{2\cdot \:1},\:x_2=\frac{-\left(-51\right)-11}{2\cdot \:1}\\ x=31,\:x=20\\ Probamos\ con\ ambos\ resultados\ y\ descartamos\ 31 (No sale)\\\ x=20$