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Algebra · real student question

Simplify the square root of (20 x^2 y^2) / z^5.

Question

Simplify

20x2y2z5\sqrt{\frac{20x^2y^2}{z^5}}

Step-by-step solution

  1. Split every factor into a perfect square times a leftover.

    20=4⋅5,z5=z4⋅z20=4\cdot 5,\qquad z^5=z^4\cdot z

    so the radicand becomes

    4⋅5⋅x2y2z4⋅z\frac{4\cdot 5\cdot x^2y^2}{z^4\cdot z}

  2. Take the square roots of the perfect squares, using absolute values where needed. u2=∣u∣\sqrt{u^2}=|u|, not uu:

    4=2,x2=∣x∣,y2=∣y∣,z4=z2\sqrt4=2,\quad\sqrt{x^2}=|x|,\quad\sqrt{y^2}=|y|,\quad\sqrt{z^4}=z^2

    No bars are needed on z2z^2 because it is already non-negative. The result so far:

    2∣x∣∣y∣5z2z\frac{2|x||y|\sqrt5}{z^2\sqrt z}

  3. Note the domain. The original requires z>0z>0 (an odd power of zz in the denominator under a square root), which is why zz itself needs no absolute value.

  4. Rationalise the denominator. Multiply numerator and denominator by z\sqrt z:

    2∣x∣∣y∣5z2z⋅zz=2∣xy∣5zz3\frac{2|x||y|\sqrt5}{z^2\sqrt z}\cdot\frac{\sqrt z}{\sqrt z}=\frac{2|xy|\sqrt{5z}}{z^3}

    since 5⋅z=5z\sqrt5\cdot\sqrt z=\sqrt{5z} and z2⋅z=z3z^2\cdot z=z^3.

  5. Verify numerically. Take x=2x=2, y=3y=3, z=4z=4: the radicand is 20⋅4⋅91024=0.703125\tfrac{20\cdot 4\cdot 9}{1024}=0.703125, whose root is 0.838530.83853. The simplified form gives 2⋅62064=12⋅4.4721464=0.83853\tfrac{2\cdot 6\sqrt{20}}{64}=\tfrac{12\cdot 4.47214}{64}=0.83853 ✓\checkmark. If all variables are assumed positive the answer may be written 2xy5zz3\tfrac{2xy\sqrt{5z}}{z^3}.

Answer

20x2y2z5=2∣xy∣5zz3(z>0)\sqrt{\frac{20x^2y^2}{z^5}}=\frac{2|xy|\sqrt{5z}}{z^3}\qquad(z>0)

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