Question

$132x2−251x+117=0 \: \: then \: \: find \: the \: possible \: vaule \: ofx$
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Answer 1

Answer:

$132x^2a^{251}x+117=0\quad :\quad x=-\sqrt[3]{\frac{39}{44a^{251}}},\:x=\frac{\sqrt[3]{39}}{2\cdot \:2^{\frac{2}{3}}\sqrt[3]{11}a^{\frac{251}{3}}}-i\frac{3^{\frac{5}{6}}\sqrt[3]{13}}{2\cdot \:2^{\frac{2}{3}}\sqrt[3]{11}a^{\frac{251}{3}}},\:$

$+\ i \frac{3\frac{5}{6} \sqrt[3]{13} }{2*2\frac{2}3}\sqrt[3]{11} a\frac{251}{3} } ; a\neq 0$

Step-by-step explanation:

$132x^2a^{251}x+117=0$

$\mathrm{Subtract\:}117\mathrm{\:from\:both\:sides}$

$132x^2a^{251}x+117-117=0-117$

$132x^2a^{251}x=-117$

$132a^{251}x^3=-117$

$\mathrm{Divide\:both\:sides\:by\:}132a^{251};\quad \:a\ne \:0$

$\frac{132a^{251}x^3}{132a^{251}}=\frac{-117}{132a^{251}};\quad \:a\ne \:0$

$x^3=-\frac{39}{44a^{251}};\quad \:a\ne \:0$

$\mathrm{For\:}x^3=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt[3]{f\left(a\right)},\:\sqrt[3]{f\left(a\right)}\frac{-1-\sqrt{3}i}{2},\:\sqrt[3]{f\left(a\right)}\frac{-1+\sqrt{3}i}{2}$

$x=\sqrt[3]{-\frac{39}{44a^{251}}},\:x=\sqrt[3]{-\frac{39}{44a^{251}}}\frac{-1+\sqrt{3}i}{2},\:x=\sqrt[3]{-\frac{39}{44a^{251}}}\frac{-1-\sqrt{3}i}{2};\quad \:a\ne \:0$

$x=-\sqrt[3]{\frac{39}{44a^{251}}},\:x=\frac{\sqrt[3]{39}}{2\cdot \:2^{\frac{2}{3}}\sqrt[3]{11}a^{\frac{251}{3}}}-i\frac{3^{\frac{5}{6}}\sqrt[3]{13}}{2\cdot \:2^{\frac{2}{3}}\sqrt[3]{11}a^{\frac{251}{3}}},\\\\\:x=\frac{\sqrt[3]{39}}{2\cdot \:2^{\frac{2}{3}}\sqrt[3]{11}a^{\frac{251}{3}}}+i\frac{3^{\frac{5}{6}}\sqrt[3]{13}}{2\cdot \:2^{\frac{2}{3}}\sqrt[3]{11}a^{\frac{251}{3}}};\quad \:a\neq 0$