How do you simplifiy $\sqrt { 1573x ^ { 7} y ^ { 2} z ^ { 4} }$ ?

Question

1786 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-09T23:09:47

$11|\x^3y|z^2sqrt(13x)$

Let's find the perfect squares in the radical. Recall that $sqrt(ab)=sqrt(a*b)$ , $sqrt(a^(b+c))=sqrt(a^b*a^c)$ , and $(a^b)^c=a^(b*c)=a^(bc)$

$sqrt(121*13*x^6*x*y^2*z^4) rarr$ 121, $x^6$ , $y^2$ , and $z^4$ are all perfect squares. They can be taken out of the radical.

$11^2=121$

$(x^3)^2=x^6$

$y^2=y^2$

$(z^2)^2=z^4$

$11|\x^3y|z^2sqrt(13x) rarr$ Absolute value bars need to be placed around the odd exponents so that they are positive (not around the even ones)

Answer 2 · 2018-04-10T00:29:59

$sqrt(1573x^7y^2z^4)=11sqrt(13)x^(7/2)|y|z^2$

$sqrt(1573x^7y^2z^4)=sqrt(13)(11^2x^7y^2z^4)^(1/2)$

$=11sqrt(13)x^(7/2)|y|z^2$ .

How do you simplifiy \sqrt { 1573x ^ { 7} y ^ { 2} z ^ { 4} }\sqrt { 1573x ^ { 7} y ^ { 2} z ^ { 4} }?