Question

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

$\displaystyle \tt 32-(2k+1)^2=(2k-3)^2\\\displaystyle \tt 32-(4k^2+4k+1)=4k^2-12k+9\\\displaystyle \tt 32-4k^2-4k-1=4k^2-12k+9\\\displaystyle \tt 31-4k^2-4k=4k^2-12k+9\\\displaystyle \tt 31-4k^2-4k-4k^2+12k-9=0\\\displaystyle \tt 22-8k^2-4k+12k=0\\\displaystyle \tt -8k^2+8k+22=0 \: \: \: \: \: | \div (-2)\\\displaystyle \tt 4k^2-4k-11=0\\\displaystyle \tt D=(-4)^2-4\cdot4\cdot(-11)=16+176=192\\\displaystyle \tt \sqrt{D}=\sqrt{192}=8\sqrt{3}$

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