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Nady [450]
3 years ago
5

Find the positive value of x in the geometric sequence 7x-2,4x+4,3x

Mathematics
1 answer:
Goryan [66]3 years ago
5 0
Hello,


Let's assume k the ratio.


 \left \{ {{3x=k*(4x+4)} \atop {4x+4=k*(7x-2)}} \right. \\\\
 \left \{ {{3x-4kx=4k} \atop {4x-7kx=-2k-4}} \right. \\\\
 \left \{ {{x(3-4k)=4k} \atop {x(4-7k)=-2k-4}} \right. \\\\
 \left \{ {{x=\dfrac{4k}{3-4k} \atop {x=\dfrac{-2k-4}{4-7k} \right. \\\\

\dfrac{4k}{3-4k}=\dfrac{-2k-4}{4-7k}\\\\

36k^2-6k-12=0\\
6k^2-k-2=0\\

\boxed{k= \dfrac{2}{3} \ or\ k= -\dfrac{1}{2}}\\\\

x=\dfrac{4*\dfrac{2}{3} }{3-4*\dfrac{2}{3} }=8\\
x=\dfrac{4*\dfrac{-1}{2} }{3-4*\dfrac{-1}{2} }=-\dfrac{2}{5}\\







\boxed{x= 8 \ or\ x= -\dfrac{2}{5}}\\\\





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<img src="https://tex.z-dn.net/?f=%5Csqrt%7B%28-81%29x%5E%7B2%7D%20%7D" id="TexFormula1" title="\sqrt{(-81)x^{2} }" alt="\sqrt{(
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Answer:

We conclude that:

\sqrt{\left(-81\right)x^2}=9ix

Step-by-step explanation:

Given the radical expression

\sqrt{\left(-81\right)x^2}

simplifying the expression

\sqrt{\left(-81\right)x^2}

Remove parentheses:  (-a) = -a

\sqrt{\left(-81\right)x^2}=\sqrt{-81x^2}

Apply radical rule:   \sqrt{-a}=\sqrt{-1}\sqrt{a},\:\quad \mathrm{\:assuming\:}a\ge 0

                 =\sqrt{-1}\sqrt{81x^2}

Apply imaginary number rule:  \sqrt{-1}=i

                 =i\sqrt{81x^2}

Apply radical rule:   \sqrt[n]{ab}=\sqrt[n]{a}\sqrt[n]{b},\:\quad \mathrm{\:assuming\:}a\ge 0,\:b\ge 0

                  =\sqrt{81}i\sqrt{x^2}

                  =9i\sqrt{x^2}

Apply radical rule:  \sqrt[n]{a^n}=a,\:\quad \mathrm{\:assuming\:}a\ge 0

                  =9ix

Therefore, we conclude that:

\sqrt{\left(-81\right)x^2}=9ix

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