Question 303313
multiply then simplify by factoring.
Use this as a cube root {{{3sqrt(n)}}} symbol
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(3rd root)of y^7 * (3rd root) of 16y^8
{{{3sqrt(y^7) * 3sqrt(16y^8)}}}
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We can combine these under single radical
{{{3sqrt(16y^8y^7)}}}
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Add exponents when you multiply
{{{3sqrt(16y^15)}}}
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Factor to reveal perfect cubes
{{{3sqrt(8*2*y^3*y^3*y^3*y^3*y^3)}}}
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Extract those perfect cubes
{{{2*y^5*3sqrt(2)}}}
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Use Rational exponents to simplify the third root of x^6
to raise on power to another power, multiply exponents
{{{(x^6)^(1/3)}}} = {{{x^(6*(1/3))}}} = {{{x^2}}}
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Find the 16th root of (-8)^16
{{{((-8)^16)^(1/16)}}} = {{{(-8)^(16*(1/16))}}} = {{{(-8)^1}}} = -8
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Rewrite the 4th root of 23 with a rational exponent.
{{{23^(1/4)}}}
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Simplify the 3rd root of -1/64
{{{3sqrt(-1/64)}}} = {{{3sqrt((-1/4)*(-1/4)*(-1/4))}}} = {{{(-1/4)}}}
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Simplify by taking roots of the numerator and denominator
(3rd root) of 1000x^14/y^3
{{{3sqrt(1000x^14/y^3)}}}
Extract cube root 1/y^3
{{{(1/y)*3sqrt(1000x^14)}}}
Extract the cube root of 1000
{{{(10/y)*3sqrt(x^14)}}}
Factor inside the radical
{{{(10/y)*3sqrt(x^3*x^3*x^3*x^3*x^2)}}}
Extract
{{{(10/y)*x^4*3sqrt(x^2)}}}
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;
Solve
The square root of 5x+29 = x+3 
{{{sqrt(5x+29) = x+3}}}
square both sides:
5x + 29 = (x+3)^2
FOIL the right side
5x + 29 = x^2 + 6x + 9
combine like terms on the right
0 = x^2 + 6x - 5x + 9 - 29
A quadratic equation
x^2 + x - 20 = 0
Factors to
(x+5)(x-4) = 0
two solutions
x = -5
and
x = +4
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Check both solution in original equation
x = -5
{{{sqrt(5)-5)+29) = -5+3}}}
{{{sqrt(-25+29) = -2}}}
{{{sqrt(4) = -2}}} not a solution
x = 4
{{{sqrt(5(4)+29) = 4+3}}}
{{{sqrt(20+29) = 7}}}
{{{sqrt(49) = 7}}}; a good solution (x=4)