Simplifying Radical Expressions: Free Response

Split the radicand into a perfect square times a remainder. Since $63 = 9 \cdot 7$,

$63x^2 = 9x^2 \cdot 7.$

Both factors on the right are non-negative for every real $x$, so the product rule applies:

$\sqrt{63x^2} = \sqrt{9x^2}\,\sqrt 7 = \sqrt{9}\,\sqrt{x^2}\,\sqrt 7 = 3|x|\sqrt 7.$

The bars are needed because $\sqrt{x^2}$ is $|x|$, not $x$: at $x = -1$ the original expression is $\sqrt{63} \approx 7.94$, a positive number, while $3(-1)\sqrt7 \approx -7.94$ would have the wrong sign.

The radicand $81x^8$ is itself a perfect square, since $81x^8 = (9x^4)^2$. Extracting it directly,

$\sqrt{81x^8} = \sqrt{(9x^4)^2} = |9x^4|.$

Now decide whether the bars change anything. $9x^4 \ge 0$ for every real $x$, because $x^4$ is an even power and $9$ is positive, so $|9x^4| = 9x^4$ outright:

$\sqrt{81x^8} = 9x^4.$

No bars are needed in the final answer, not because the identity $\sqrt{a^2}=|a|$ stopped applying, but because the quantity inside those bars was already non-negative, so the bars had nothing to fix.

Halve the exponent inside the radical to find the exponent that would emerge: $14 \div 2 = 7$, so $\sqrt{x^{14}} = |x^7|$ before any bars are removed. The question of whether those bars survive comes down entirely to the parity of that emerging exponent, $7$.

An exponent that comes out even always leaves a non-negative quantity ($x^{\text{even}} \ge 0$ for every real $x$), so its bars can be dropped, as happened in part B. An exponent that comes out odd leaves a quantity that can still be negative, since an odd power carries the sign of $x$, so the bars are load-bearing there. Here the emerging exponent is $7$, which is odd, so the bars stay:

$\sqrt{x^{14}} = |x^7| = |x|^7.$

Convert each factor to $i$ form first, since neither radicand is non-negative.

$\sqrt{-9} = 3i, \qquad \sqrt{-49} = 7i$

Multiply the two results.

$\sqrt{-9}\cdot\sqrt{-49} = (3i)(7i) = 21i^2 = -21$

Multiply the two radicands together first, then take the square root of the single non-negative result.

$\sqrt{(-9)(-49)} = \sqrt{441} = 21$

Compare this to part A, which gave $-21$. The two results are not equal; they are negatives of each other. Since $a=-9$ and $b=-49$ are both negative, the hypothesis of the product rule fails, and the rule genuinely does not apply here: computing the product first and computing each root separately land on the two different square roots of $441$.

Convert the negative radicand to $i$ form and evaluate both sides.

$\sqrt{-9}\cdot\sqrt{64} = (3i)(8) = 24i$

$\sqrt{(-9)(64)} = \sqrt{-576} = 24i$

The two agree. With only one negative factor there is no competition between two candidate square roots the way there was in part B: the product $ab$ is negative either way, and its unique square root in $i$ form is reached the same way from either side.

Putting parts A through C together, the pattern is settled. The rule is safe whenever at most one of the two radicands is negative, and it is the case of two negative radicands, and only that case, where it breaks.

The side length of a square field is the square root of its area. Split $245$ into a perfect square times a remainder: $245 = 49 \cdot 5$.

$\sqrt{245} = \sqrt{49}\,\sqrt5 = 7\sqrt5 \text{ meters}$

Write the second area as $4$ times the first, and split the square root across that product.

$\sqrt{980} = \sqrt{4 \cdot 245} = \sqrt4\,\sqrt{245} = 2\sqrt{245}$

Now substitute the simplified form from part A instead of expanding $245$ again.

$2\sqrt{245} = 2(7\sqrt5) = 14\sqrt5 \text{ meters}$

The quotient rule lets a single fraction under one radical sign replace two separate radicals.

$\frac{\sqrt{500}}{\sqrt5} = \sqrt{\frac{500}{5}} = \sqrt{100} = 10 \text{ meters}$

The shared hypothesis is a non-negativity condition on both radicands, with the quotient rule adding a stricter condition on its denominator: not just $\ge 0$, but $> 0$, since dividing by zero is never allowed.

$a \ge 0, \qquad b \ge 0 \quad (\text{and } b > 0 \text{ for the quotient rule})$

In every part of this question, the quantities standing in for $a$ and $b$, an area, a ratio of two areas, an area and a length, were physical measurements, and a physical area or length is never negative by its nature, which secures the $\ge 0$ half of the hypothesis everywhere.

Part C needs the stronger condition too, since its denominator is built from this problem's field measurements, and any length used to describe a real field is not just non-negative but strictly positive: a square of side $0$ would have no area to describe. So the quotient rule's stricter requirement is met here for the same physical reason, not merely inherited from the weaker non-negativity of an area.

That is exactly why none of parts A through C could ever have landed on the sign failure that shows up only when both radicands are negative, or on a division by a radicand of zero: a negative or zero radicand never enters a problem built entirely from areas and lengths, so the hypotheses these two rules require are satisfied automatically, without having to be checked case by case.

Look for a perfect-square factor of $200$. Since $200 = 100 \cdot 2$ and $100$ is a perfect square, condition 1 (no perfect $n$-th power factor left under the radical) is violated as written. Extract it:

$\sqrt{200} = \sqrt{100}\,\sqrt2 = 10\sqrt2$

Nothing further comes out: $2$ has no perfect-square factor beyond $1$.

Condition 3 requires the index and the exponents under the radical to share no common factor. Here the index is $6$ and the radicand's exponent is $4$, and $\gcd(6,4) = 2$, so condition 3 is violated as written.

Convert to a rational exponent and reduce the fraction, which is exactly what canceling that common factor means:

$\sqrt[6]{x^4} = x^{4/6} = x^{2/3} = \sqrt[3]{x^2}$

The index and the exponent of the reduced form, $3$ and $2$, share no common factor, so the result is in simplest radical form.

Evaluate the expression directly. Since $32 = 2^5$ is a perfect fifth power,

$\sqrt[5]{32} = \sqrt[5]{2^5} = 2.$

So $\sqrt[5]{32}$ does simplify all the way down to a whole number with no radical sign left at all, which means the classmate's conclusion, that the original expression was not in simplest form, is correct.

The classmate's stated reason is still wrong, though. None of the three conditions this lesson defines says a simplified answer can never carry a radical sign; condition 1 only bans a perfect $n$-th power factor from remaining inside. An expression like $\sqrt2$ is already in simplest radical form and keeps its radical sign forever, because $2$ has no perfect-square factor to extract. What actually convicted $\sqrt[5]{32}$ was that its entire radicand, $32$, is itself a perfect fifth power, which condition 1 forbids; the presence of a radical sign by itself was never the issue.

Raise each value to the sixth power first, then take the sixth root.

At $x=-3$: $(-3)^6 = 729$, and $\sqrt[6]{729} = 3$, since $3^6 = 729$.

At $x=5$: $5^6 = 15625$, and $\sqrt[6]{15625} = 5$, since $5^6 = 15625$.

Both outputs are the non-negative version of the input: $3$ from $-3$, and $5$ from $5$ itself. That is exactly the absolute value, so

$\sqrt[6]{x^6} = |x| \quad \text{for every real } x.$

By part A, $\sqrt[6]{x^6} = |x|$, so the equation to solve is $|x| = -x$.

First direction: if $x \le 0$, then by the definition of absolute value $|x| = -x$ (this is exactly the piece of the definition that applies for a non-positive input, and it includes $x=0$ since $|0|=0=-0$). So every $x \le 0$ satisfies the equation.

Second direction: if $|x| = -x$, then since $|x| \ge 0$ always, $-x \ge 0$, which forces $x \le 0$. So no value outside $x \le 0$ can satisfy the equation.

Both directions agree, which is what "exactly" requires:

$\sqrt[6]{x^6} = -x \iff x \le 0.$

By the same reasoning as part B, $\sqrt[6]{x^6} = x$ becomes $|x| = x$, which holds exactly when $x \ge 0$. Part B already established that $\sqrt[6]{x^6} = -x$ holds exactly when $x \le 0$.

A number satisfying both equations at once must satisfy both conditions at once: $x \ge 0$ and $x \le 0$. The only real number satisfying both inequalities simultaneously is

$x = 0.$

Any nonzero $x$ is either strictly positive or strictly negative, which fails one of the two inequalities, so no other real number can satisfy both equations.