MATH SOLVE

5 months ago

Q:
# I know the answers to these questions, I just can't seem to figure out how to find the answers. I'm not too good at math...Simplify the radical exprssion[tex] \sqrt{14q} * 2 \sqrt{4q} [/tex]Answer is [tex]4q \sqrt{14} [/tex]Simplify the radical expression (There's supposed to be a big suare root sign over the whole fraction) [tex] \frac{ \sqrt{63x^15y^9} }{7xy^11} [/tex] (^15 and ^11)Answer is [tex] \frac{3x^7}{y} [/tex]Simplify the radical expression[tex]2 \sqrt{6} + 3 \sqrt{96} [/tex]Answer is [tex]14 \sqrt{6} [/tex]

Accepted Solution

A:

[tex]\bf \sqrt{14q}\cdot 2\sqrt{4q}\implies \sqrt{14q}\cdot 2\sqrt{2^2q}\implies \sqrt{14q}\cdot 2(2)\sqrt{q}
\\\\\\
\sqrt{14q}\cdot 4\sqrt{q}
\implies
4\sqrt{14q}\cdot \sqrt{q}\implies 4\sqrt{14q\cdot q}\implies 4\sqrt{14q^2}
\\\\\\
4q\sqrt{14}[/tex]

now, for the next one, bear in mind that, if you move a "factor" from the bottom to the top, the exponent changes sign, and if you move it from the top to the bottom, also changes sign, so, let's see,

[tex]\bf \left.\qquad \qquad \right.\textit{negative exponents}\\\\ a^{-{ n}} \implies \cfrac{1}{a^{ n}} \qquad \qquad \cfrac{1}{a^{ n}}\implies a^{-{ n}} \qquad \qquad a^{{{ n}}}\implies \cfrac{1}{a^{-{{ n}}}}\\\\ -------------------------------\\\\ \sqrt{\cfrac{63x^{15}y^9}{7xy^{11}}}\implies \sqrt{\cfrac{63}{7}\cdot \cfrac{x^{15}y^9}{xy^{11}}}\implies \sqrt{ \cfrac{9x^{15}y^9}{xy^{11}}}\implies \sqrt{ \cfrac{9x^{15}x^{-1}}{y^{11}y^{-9}}}[/tex]

[tex]\bf \sqrt{ \cfrac{9x^{15-1}}{y^{11-9}}}\implies \sqrt{\cfrac{9x^{14}}{y^2}}\implies \sqrt{\cfrac{9x^{7\cdot 2}}{y^2}}\implies \sqrt{\cfrac{3^2(x^7)^2}{y^2}}\implies \cfrac{3x^7}{y}\\\\ -------------------------------[/tex]

[tex]\bf 2\sqrt{6}+3\sqrt{96}\qquad \begin{cases} 96=2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 3\\ \qquad 2^2\cdot 2^2\cdot 2\cdot 3\\ \qquad (2\cdot 2)^2\cdot 2\cdot 3\\ \qquad 4^2\cdot 6 \end{cases}\implies 2\sqrt{6}+3\sqrt{4^2\cdot 6} \\\\\\ 2\sqrt{6}+3(4)\sqrt{6}\implies 2\sqrt{6}+12\sqrt{6}\implies 14\sqrt{6}[/tex]

now, for the next one, bear in mind that, if you move a "factor" from the bottom to the top, the exponent changes sign, and if you move it from the top to the bottom, also changes sign, so, let's see,

[tex]\bf \left.\qquad \qquad \right.\textit{negative exponents}\\\\ a^{-{ n}} \implies \cfrac{1}{a^{ n}} \qquad \qquad \cfrac{1}{a^{ n}}\implies a^{-{ n}} \qquad \qquad a^{{{ n}}}\implies \cfrac{1}{a^{-{{ n}}}}\\\\ -------------------------------\\\\ \sqrt{\cfrac{63x^{15}y^9}{7xy^{11}}}\implies \sqrt{\cfrac{63}{7}\cdot \cfrac{x^{15}y^9}{xy^{11}}}\implies \sqrt{ \cfrac{9x^{15}y^9}{xy^{11}}}\implies \sqrt{ \cfrac{9x^{15}x^{-1}}{y^{11}y^{-9}}}[/tex]

[tex]\bf \sqrt{ \cfrac{9x^{15-1}}{y^{11-9}}}\implies \sqrt{\cfrac{9x^{14}}{y^2}}\implies \sqrt{\cfrac{9x^{7\cdot 2}}{y^2}}\implies \sqrt{\cfrac{3^2(x^7)^2}{y^2}}\implies \cfrac{3x^7}{y}\\\\ -------------------------------[/tex]

[tex]\bf 2\sqrt{6}+3\sqrt{96}\qquad \begin{cases} 96=2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 3\\ \qquad 2^2\cdot 2^2\cdot 2\cdot 3\\ \qquad (2\cdot 2)^2\cdot 2\cdot 3\\ \qquad 4^2\cdot 6 \end{cases}\implies 2\sqrt{6}+3\sqrt{4^2\cdot 6} \\\\\\ 2\sqrt{6}+3(4)\sqrt{6}\implies 2\sqrt{6}+12\sqrt{6}\implies 14\sqrt{6}[/tex]