Question

cube root x square + 2 equal to 3​

Answer 1

Answer:

$\sqrt[3]{x^2}+2=3\quad :\quad x=1,\:x=-1$

Step-by-step explanation:

$\sqrt[3]{x^2}+2=3$

$\mathrm{Subtract\:}2\mathrm{\:from\:both\:sides}$

$\sqrt[3]{x^2}+2-2=3-2$

$\mathrm{Simplify}$

$\sqrt[3]{x^2}=1$

$\mathrm{Take\:both\:sides\:of\:the\:equation\:to\:the\:power\:of\:}3$

$\left(\sqrt[3]{x^2}\right)^3=1^3$

$\mathrm{Expand\:}\left(\sqrt[3]{x^2}\right)^3$

$\left(\sqrt[3]{x^2}\right)^3$

$\sqrt[n]{a}=a^{\frac{1}{n}}$

$=\left(x^2^{\frac{1}{3}}\right)^3$

$\mathrm{Apply\:exponent\:rule}:\quad \left(a^b\right)^c=a^{bc}$

$=x^2^{\frac{1}{3}\cdot \:3}$

$=x^2$

$\mathrm{Expand\:}1^3:\quad 1$

$x^2=1$

$\mathrm{Solve\:}\:x^2=1$

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

$x=\sqrt{1},\:x=-\sqrt{1}$

$x=1,\:x=-1$

$\mathrm{Verify\:Solutions}$

$\mathrm{Check\:the\:solutions\:by\:plugging\:them\:into\:}\sqrt[3]{x^2}+2=3$ $\mathrm{Remove\:the\:ones\:that\:don't\:agree\:with\:the\:equation.}$

$\mathrm{Plug}\:x=1:\quad \mathrm{True}$

$\mathrm{Plug}\:x=-1:\quad \mathrm{True}$

$\mathrm{The\:solution\:is}$

$x=1,\:x=-1$