Question

Os números K -3,K + 7,6K - 3 nessa ordem,uma P.G.Calcule K​

Answer 1

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$\sf a_2^2=a_1\cdot a_3\\\sf(k+7)^2=(k-3)\cdot(6k-3)\\\sf k^2+14k+49=6k^2-3k-18k+9\\\sf k^2-6k^2+14k+3k+18k+49-9=0\\\sf-5k^2+35k+40=0\div(-5)\\\sf k^2-7k-8=0\\\sf\Delta=b^2-4ac\\\sf\Delta=(-7)^2-4\cdot1\cdot(-8)\\\sf\Delta=49+32\\\sf\Delta=81\\\sf k=\dfrac{-b\pm\sqrt{\Delta}}{2a}\\\sf k=\dfrac{-(-7)\pm\sqrt{81}}{2\cdot1}\\\sf k=\dfrac{7\pm9}{2}\begin{cases}\sf k_1=\dfrac{7+9}{2}=\dfrac{16}{2}=8\\\sf k_2=\dfrac{7-9}{2}=-\dfrac{2}{2}=-1\end{cases}$