A-b=10 a*b=7 (a+b)^2=?

Составим систему: [latex]\left \{ {{a-b=10} \atop {a*b=7}} \right. \Rightarrow \left \{ {{a=10+b} \atop {b(10+b)=7}} \right. \Rightarrow \left \{ {{a=10+b} \atop {10b+b^2=7}} \right.[/latex] Решим: [latex]b^2+10b-7=0[/latex] [latex] \sqrt{D}= \sqrt{100+28}= \sqrt{128}=4\sqrt{8}=8\sqrt{2} [/latex] [latex]b_{1,2}= \frac{-10\pm8 \sqrt{2}}{2}=- \frac{2(5\pm4 \sqrt{2})}{2}=-5\pm4 \sqrt{2} [/latex] Найдем a: [latex]a_1=10-5+4\sqrt{2} =5+4\sqrt{2}=5+4\sqrt{2}[/latex] [latex]a_2=10-5-4\sqrt{2}=5-4 \sqrt{2}=5-4\sqrt{2} [/latex] Отсюда получаем: [latex](a_1+b_1)^2=[(5+4 \sqrt{2})+(-5+4 \sqrt{2})]^2=(8 \sqrt{2})^2=64*2=128 [/latex] [latex](a_2+b_2)^2=[(5-4 \sqrt{2})+(-5-4 \sqrt{2})]^2=(-8 \sqrt{2})^2=64*2=128 [/latex] Отсюда следует: [latex](a+b)^2=128[/latex]