Radicals and Rational Exponents: Chapter Test

Solution

Correct answer: A

You chose B: Applies the odd-root sign convention to an even index, as if a fourth root could return a negative number.

You chose C: Assumes that an even root failing to be real for a negative radicand means no root at any index can succeed, overlooking that the index $5$ is odd.

You chose D: Forgets that an even-indexed root of a negative radicand has no real value at all.

Solution

Correct answer: B

You chose A: Flips the sign of the value instead of taking a reciprocal.

You chose C: Drops the negative exponent instead of reciprocating.

You chose D: Reciprocates correctly but then also flips the sign of the result.

Solution

Correct answer: C

You chose A: Drops the absolute value bars, as if $x$ were known to be nonnegative.

You chose B: Pulls the factor $25$ out of the radical without taking its square root.

You chose D: Confuses the single value of a radical with the two solutions of an equation like $x^2 = 75$.

Solution

Correct answer: D

You chose A: Filters candidates by whether the radicand is nonnegative rather than substituting into the original equation; both $6(1)+3$ and $6(13)+3$ are positive, so this check wrongly passes both.

You chose B: Checks against the sign-flipped twin equation $\sqrt{6x+3} = -(x-4)$ instead of the original.

You chose C: Squares the equation as $6x+3 = x^2 + 16$, dropping the cross term from $(x-4)^2$.

Solution

Correct answer: B

You chose A: Multiplies top and bottom by $\sqrt5$ but forgets to divide by the resulting denominator, $5$.

You chose C: Drops the numerator's coefficient $3$ while rationalizing.

You chose D: Squares the denominator a second time, using $25$ instead of $5$.

Solution

Correct answer: D

You chose A: Assumes an even root always undoes an even power exactly, ignoring the absolute value it must return.

You chose B: Flips the inequality, confusing when $|x-2| = -(x-2)$ with when $|x-2| = x-2$.

You chose C: Believes bars are unconditionally required and that the radical can never equal $x-2$ itself.

Solution

Correct answer: A

You chose B: Adds the radicands directly, as if $\sqrt a + \sqrt b = \sqrt{a+b}$.

You chose C: Judges the two radicals unlike before simplifying either one.

You chose D: Multiplies the coefficients of the like radicals instead of adding them.

Solution

Correct answer: C

You chose A: Combines the two cross-term coefficients by subtracting instead of adding, $-12-2$ rather than $-12+2$.

You chose B: Drops the negative sign when multiplying $\sqrt6\cdot(-3\sqrt6)$, treating it as $+18$ instead of $-18$.

You chose D: Multiplies only the first and last terms of the two binomials, omitting both cross terms.

Solution

Correct answer: D

You chose A: Multiplies the radicands directly as $\sqrt{(-20)(-5)} = \sqrt{100} = 10$, applying the product rule where both factors are negative.

You chose B: Converts only one radical to $i$-form, leaving a single factor of $i$.

You chose C: Adds the radicands instead of multiplying them, then converts the negative sum to $i$-form.

Solution

Correct answer: B

You chose A: Multiplies the indices $4$ and $2$ instead of adding the exponents $\tfrac14$ and $\tfrac12$.

You chose C: Adds the indices, $4+2=6$, instead of finding a common denominator for the exponents.

You chose D: Treats $\sqrt5$ as $5^1$ instead of $5^{1/2}$, so the exponents combine to $\tfrac54$.

Solution

Correct answer: C

You chose A: Drops the negative sign on $b^{-1/2}$ before applying the outer power, treating it as $b^{1/2}$, so $\left(b^{1/2}\right)^6=b^3$ lands in the numerator instead of $b^{-3}$ in the denominator.

You chose B: Subtracts the exponent $\tfrac16$ from $\tfrac23$ without flipping the sign of the negative exponent it is dividing by.

You chose D: Raises only the $a$-term to the outer power $6$ and leaves $b^{-1/2}$ unchanged.

Solution

Correct answer: A

You chose B: Uses $a^2+b^2d$ instead of $a^2-b^2d$ in the denominator, as if rationalizing a complex denominator instead of a radical one.

You chose C: Multiplies the denominator by its conjugate but leaves the numerator unchanged, changing the value of the expression.

You chose D: Flips the sign of the correctly computed result at the very end.

Solution

Correct answer: D

You chose A: Raises both sides to the 4th power without checking that the principal root can never equal a negative number, getting $2x-5=81$.

You chose B: Incorrectly keeps $(-3)^4$ negative (getting $-81$), then isolates $x$ by subtracting $5$ instead of adding it, computing $2x=-81-5=-86$.

You chose C: Ignores the index $4$ and treats the radical as an ordinary square root, squaring both sides to get $2x-5=9$.

Solution

Correct answer: B

You chose A: Uses $a^2 - bd$ instead of $a^2 - b^2d$, forgetting to square the coefficient $3$.

You chose C: Adds instead of subtracts, as if this were a complex-conjugate product.

You chose D: Computes the square of the sum instead of recognizing the conjugate pair, doubling the cross term instead of canceling it.

Solution

Correct answer: A

You chose B: Drops the absolute value bars, assuming $x > 0$.

You chose C: Extracts $\sqrt{4x^2}$ as $4|x|$ instead of $2|x|$, forgetting to take the square root of the coefficient $4$.

You chose D: Introduces an extra factor of $x$ in the denominator, as if squaring it a second time while clearing the radical.

Solution

Correct answer: C

You chose A: Discards the negative candidate $x=-2$ on sight, without checking it in the original equation.

You chose B: Makes an arithmetic slip checking $x=-1$, computing $\sqrt4$ as $-2$ and rejecting a genuine solution.

You chose D: Misreads the right side of the equation as $-1$, which would have no solution since $\sqrt{3x+7} \ge \sqrt{x+2}$ throughout the domain.

Solution

Correct answer: C

You chose A: Pushes the base through the power law despite the base being negative, exactly the move the exponent laws do not license.

You chose B: Concludes the expression is undefined merely because the unreduced and reduced routes disagree, without checking whether the reduced fraction's denominator is odd, the actual test this chapter's rational-exponents materials established for whether a negative base's rational power is real.

You chose D: Believes a negative base can be freely manipulated once the numerator happens to be even.

Solution

Correct answer: D

You chose A: Uses the sign pattern for $(x+y)(x^2-xy+y^2)$ instead of $(x-y)(x^2+xy+y^2)$, flipping the middle sign.

You chose B: Treats $(\sqrt[3]3)^2$ as the rational number $3$, as if it were a square root, and believes $(2-\sqrt[3]3)(2+\sqrt[3]3)=4-3=1$.

You chose C: Multiplies by the two-term sign flip $2+\sqrt[3]3$ and wrongly assumes $(2-\sqrt[3]3)(2+\sqrt[3]3)$ still equals $5$, as if it produced the same rational denominator as the correct three-term multiplier, so the $5$s cancel and only $2+\sqrt[3]3$ is left.

Solution

Correct answer: C

You chose A: Assumes the inner cube root must be nonnegative, as it would be for an even index, and discards the valid negative branch.

You chose B: Treats raising both sides to the reciprocal power $\tfrac32$ as multiplying by $\tfrac32$ instead, computing $3x-2=4\cdot\tfrac32=6$ rather than $3x-2=4^{3/2}=8$.

You chose D: Takes only the positive branch $u=2$, then isolates $x$ with a sign slip, computing $3x=2-8$ instead of $3x=8+2$.

Solution

Correct answer: A

You chose B: Computes the conjugate-pair simplification correctly but ignores that each individual square root requires a nonnegative radicand.

You chose C: Wrongly generalizes from the single boundary value $x=2$, where $(x+5)-(2-x)=7$, to the whole domain.

You chose D: Enforces only the first radicand's condition, $x+5\ge0$, and forgets the second, $2-x\ge0$.