Question

#100 points•


✨✨(Q) X²- 6√x + 0 ✨✨


⛓please solve this ⛓

❤Need quality answers ❤



☺follow me guys ☺

Answer 1

$\underline{\texttt{Solution :}}$

$\text{The given equation is}$

$\mathrm{x^{2}-6\sqrt{x}=0}$

$\to \mathrm{x^{2}=6\sqrt{x}}$

$\to \mathrm{(x^{2})^{2}=(6\sqrt{x})^{2}}$

$\to \mathrm{x^{4}=36x}$

$\to \mathrm{x^{4}-36x=0}$

$\to \mathrm{x(x^{3}-36)=0}$

$\mathrm{So,\:we\:have\:x = 0\:and\:x^{3}-36=0}$

$\mathrm{Now,\:x^{3}=36}$

$\to \mathrm{x=36^{\frac{1}{3}}}$

$\to \mathrm{x = {(6^{2})}^{\frac{1}{3}}}$

$\to \mathrm{x = 6^{\frac{2}{3}}}$

$\text{Now, the equation}$

$\mathrm{x^{3}-36=0\:can\:be\:written\:as}$

$\to \mathrm{(x-6^{\frac{2}{3}})(x^{2}+6^{\frac{2}{3}}\:x+6^{\frac{4}{3}})=0}$

$\implies \mathrm{x=6^{\frac{2}{3}}\:,\:x^{2}+6^{\frac{2}{3}}\:x+6^{\frac{4}{3}}=0}$

$\text{From the 2nd equation, we get}$

$\mathrm{x=\frac{-6^{\frac{2}{3}}\pm \sqrt{6^{\frac{4}{3}}-4*6^{\frac{4}{3}}}}{2}}$

$\mathrm{=\frac{-6^{\frac{2}{3}}\pm \sqrt{-3*6^{\frac{4}{3}}}}{2}}$

$\mathrm{=\frac{-6^{\frac{2}{3}}\pm i6^{\frac{2}{3}}\sqrt{3}}{2}}$

$\:\:\:\:\:\mathrm{where\:i=\sqrt{-1}}$

$\mathrm{=6^{\frac{2}{3}}(\frac{-1\pm i\sqrt{3}}{2})}$

$\therefore \text{the complete solution be}$

$\mathrm{x=0\:,\:6^{\frac{2}{3}}\:,6^{\frac{2}{3}}(\frac{-1\pm i\sqrt{3}}{2})}$

$\to \boxed{\mathrm{x=0\:,\:\sqrt[3]{36}\:,\:\sqrt[3]{36}(\frac{-1\pm i\sqrt{3}}{2})}}$

$\underline{\texttt{To find a single root :}}$

$\mathrm{Now,\:x^{2}-6\sqrt{x}=0}$

$\to \mathrm{x^{2}=6x^{\frac{1}{2}}}$

$\to \mathrm{log(x^{2})=log(6x^{\frac{1}{2}})}$

$\to \mathrm{2logx=log6+log x^{\frac{1}{2}}}$

$\to \mathrm{2logx=log6+\frac{1}{2}logx}$

$\to \mathrm{(2-\frac{1}{2})logx=log6}$

$\to \mathrm{\frac{3}{2}logx=log6}$

$\to \mathrm{logx=\frac{2}{3}log6}$

$\to \mathrm{logx=log6^{\frac{2}{3}}}$

$\to \mathrm{x=6^{\frac{2}{3}}}$

$\to \mathrm{x=(6^{2})^{\frac{1}{3}}}$

$\to \boxed{\mathrm{x= \sqrt[3]{36}}}$