Question

if x2+1/25x2=43/8. Find x3+1/125x3​

Answer 1

Answer:

$x^2+\frac{1}{25}x^2=\frac{43}{8},\:x^3+\frac{1}{125}x^3\quad :\quad x=\frac{5\sqrt{559}}{52},\:x=-\frac{5\sqrt{559}}{52}$

$\left(\mathrm{Decimal}:\quad x=2.27338\dots ,\:x=-2.27338\dots \right)$

Step-by-step explanation:

$x^2+\frac{1}{25}x^2=\frac{43}{8},\:x^3+\frac{1}{125}x^3$

$\mathrm{Multiply\:both\:sides\:by\:}25$

$x^2\cdot \:25+\frac{1}{25}x^2\cdot \:25=\frac{43}{8}\cdot \:25$

$25x^2+x^2=\frac{1075}{8}$

$26x^2=\frac{1075}{8}$

$\mathrm{Divide\:both\:sides\:by\:}26$

$\frac{26x^2}{26}=\frac{\frac{1075}{8}}{26}$

$x^2=\frac{1075}{208}$

$\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}$

$x=\sqrt{\frac{1075}{208}},\:x=-\sqrt{\frac{1075}{208}}$

$=\frac{\sqrt{1075}}{\sqrt{208}}$

$=\frac{\sqrt{1075}}{4\sqrt{13}}$

$=\frac{5\sqrt{43}}{4\sqrt{13}}$

$=\frac{5\sqrt{43}\sqrt{13}}{4\sqrt{13}\sqrt{13}}$

$\sqrt{43}\sqrt{13}=\sqrt{43\cdot \:13}$

$=5\sqrt{43\cdot \:13}$

$\mathrm{Multiply\:the\:numbers:}\:43\cdot \:13=559$

$=5\sqrt{559}$

$4\sqrt{13}\sqrt{13}$

$\sqrt{13}\sqrt{13}=13$

$=4\cdot \:13$

$\mathrm{Multiply\:the\:numbers:}\:4\cdot \:13=52$

$=52$

$=\frac{5\sqrt{559}}{52}$

$x=\frac{5\sqrt{559}}{52},\:x=-\frac{5\sqrt{559}}{52}$

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