Question

Encontre o valor de x.

Answer 1

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$\Huge\boxed{\sf{\underline{Relac_{\!\!,}\tilde{a}o~entre~as~cordas}}}\\\huge\boxed{\boxed{\boxed{\boxed{\boxed{\sf PA\cdot PB= PC\cdot PD}}}}}$

$\Huge\boxed{\sf{\underline{Relac_{\!\!,}\tilde{a}o~entre~secantes}}}\\\huge\boxed{\boxed{\boxed{\boxed{\sf PA\cdot PB=PC\cdot PD}}}}$

$\Large\boxed{\sf{\underline{Relac_{\!\!,}\tilde{a}o~entre~secante~e~tangente}}}\\\huge\boxed{\boxed{\boxed{\boxed{\boxed{\sf PT^2=PA\cdot PB}}}}}$

$\tt a)~\sf 3\cdot(x-3)=6\cdot9\\\sf 3x-9=54\\\sf3x=54+9\\\sf 3x=63\\\sf x=\dfrac{63}{3}\\\huge\boxed{\boxed{\boxed{\boxed{\boxed{\sf x=21\checkmark}}}}}$

$\tt b)~\sf 6\cdot x= 3\cdot15\\\sf x=\dfrac{\diagup\!\!\!3^1\cdot15}{\diagup\!\!\!6_2}\\\huge\boxed{\boxed{\boxed{\boxed{\boxed{\sf x=\dfrac{15}{2}}}}}}$

$\tt c)~\sf 12^2=x\cdot(x+10)\\\sf x^2+10x=144\\\sf x^2+10x-144=0\\\sf\Delta=b^2-4ac\\\sf\Delta=10^2-4\cdot1\cdot(-144)\\\sf\Delta=100+576\\\sf\Delta=676\\\sf x=\dfrac{-b\pm\sqrt{\Delta}}{2a}\\\sf x=\dfrac{-10\pm\sqrt{676}}{2\cdot1}\\\sf x=\dfrac{-10\pm26}{2}\begin{cases}\sf x_1=\dfrac{-10+26}{2}=\dfrac{16}{2}=8\checkmark\\\sf x_2=\dfrac{-10-26}{2}=-\dfrac{36}{2}=\diagdown\!\!\!\!\!\!-\diagup\!\!\!\!\!\!\!\!\!\!18\end{cases}$