Question

calcule o valor de h na equação x^2-2×(h+1)×x+h-10=0,com incógnita X,de modo que a soma dos inversos das soluções Seja 1/3.

Answer 1

$\mathrm{x^2-2(h+1)x+h-10=0\ \ \bigg\| \ \ \dfrac{1}{x_1}+\dfrac{1}{x_2}=\dfrac{1}{3}}\\\\ \mathrm{a=1\ \| \ b=-2(h+1)\ \| \ c=h-10}\\\\ \textbf{Atrav\'es da f\'ormula quadr\'atica, teremos que:}\\\\ \mathrm{x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}}\\\\ \mathrm{x=\dfrac{-\big(-2(h+1)\big)\pm\sqrt{\big(-2(h+1)\big)^2-4.1(h-10)}}{2.1}}\\\\ \mathrm{x=\dfrac{2(h+1)\pm\sqrt{4(h^2+2h+1)+4(10-h)}}{2}}\\\\ \mathrm{x=\dfrac{2(h+1)\pm\sqrt{4(h^2+2h+1-h+10)}}{2}}$

$\mathrm{x=\dfrac{2(h+1)\pm2\sqrt{h^2+h+11}}{2}=(h+1)\pm\sqrt{h^2+h+11}}\\\\ \mathrm{x_1=(h+1)+\sqrt{h^2+h+11}\ \ \|\ \ x_2=(h+1)-\sqrt{h^2+h+11}}\\\\ \mathrm{\dfrac{1}{(h+1)+\sqrt{h^2+h+11}}+\dfrac{1}{(h+1)-\sqrt{h^2+h+11}}=\dfrac{1}{3}}\\\\\\ \mathrm{\dfrac{(h+1)-\sqrt{h^2+h+11}+(h+1)+\sqrt{h^2+h+11}}{\big((h+1)+\sqrt{h^2+h+11}\big)\big((h+1)-\sqrt{h^2+h+11}\big)}=\dfrac{1}{3}}\\\\\\ \mathrm{\dfrac{2(h+1)}{(h+1)^2-(\sqrt{h^2+h+11})^2}=\dfrac{1}{3}}$

$\mathrm{\dfrac{2(h+1)}{h^2+2h+1-h^2-h-11}=\dfrac{1}{3}\ \to\ \dfrac{2(h+1)}{h-10}=\dfrac{1}{3}}\\\\\\ \mathrm{6(h+1)=h-10\ \to\ 6h+6=h-10}\\\\ \mathrm{6h-h=-10-6\ \to\ 5h=-16\ \to\ \boxed{\mathbf{h=\dfrac{-16}{5}}}}$