Question

32-(2к+1)²=(2к-3)² решите уравнение даю 10 баллов
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Answer 1

$displaystyle tt 32-(2k+1)^2=(2k-3)^2\displaystyle tt 32-(4k^2+4k+1)=4k^2-12k+9\displaystyle tt 32-4k^2-4k-1=4k^2-12k+9\displaystyle tt 31-4k^2-4k=4k^2-12k+9\displaystyle tt 31-4k^2-4k-4k^2+12k-9=0\displaystyle tt 22-8k^2-4k+12k=0\displaystyle tt -8k^2+8k+22=0 : : : : : | div (-2)\displaystyle tt 4k^2-4k-11=0\displaystyle tt D=(-4)^2-4cdot4cdot(-11)=16+176=192\displaystyle tt sqrt{D}=sqrt{192}=8sqrt{3}$

$displaystyle tt boxed{bold{x_1=frac{4+8sqrt{3}}{2cdot4}=frac{4(1+2sqrt{3})}{8}=frac{1+2sqrt{3}}{2}}}\displaystyle tt boxed{bold{x_2=frac{4-8sqrt{3}}{2cdot4}=frac{4(1-2sqrt{3})}{8}=frac{1-2sqrt{3}}{2}}}$