Question

38)Resolvendo a equação x2 - (a + b)x + ab = 0, obtém-se duas raízes. Sabendo-se que essas raízes são as diagonais de um losango, determine a área desse losango.

Answer 1

$\mathrm{x^2-(a+b)x+ab=0}\\\\ \mathrm{x=\dfrac{-[-(a+b)]\pm\sqrt{[-(a+b)]^2-4.1.ab}}{2.1}=}\\\\ \mathrm{=\dfrac{a+b\pm\sqrt{(a+b)^2-4ab}}{2}=\dfrac{a+b\pm\sqrt{a^2+2ab+b^2-4ab}}{2}=}\\\\ \mathrm{=\dfrac{a+b\pm\sqrt{a^2-2ab+b^2}}{2}=\dfrac{a+b\pm\sqrt{(a-b)^2}}{2}=\dfrac{a+b\pm(a-b)}{2}}\\\\ \mathrm{x_1=\dfrac{a+b+(a-b)}{2}=\dfrac{a+b+a-b}{2}=\dfrac{2a}{2}=a\ \to\ \boxed{\mathrm{x_1=a}}}\\\\ \mathrm{x_2=\dfrac{a+b-(a-b)}{2}=\dfrac{a+b-a+b}{2}=\dfrac{2b}{2}=b\ \to\ \boxed{\mathrm{x_2=b}}}$

$\mathrm{\Rightarrow Diagonais\ do\ losango=\{a,b\}}\\\\ \mathrm{S_{losango}=\dfrac{D.d}{2}=\dfrac{a.b}{2}\ \to\ \boxed{\boxed{\mathbf{S_{losango}=\dfrac{ab}{2}}}}}$