Solving Radical Equations Practice

The radical is already isolated; square both complete sides, solve, then substitute back.

Square both sides to get \(2x+7=25\), so \(x=9\). The original check gives \(\sqrt{25}=5\), so the candidate is valid.

The radical is already isolated; square both complete sides, solve, then substitute back.

Square both sides to get \(3x+4=16\), so \(x=4\). The original check gives \(\sqrt{16}=4\), so the candidate is valid.

The radical is already isolated; square both complete sides, solve, then substitute back.

Square both sides to get \(5x-4=36\), so \(x=8\). The original check gives \(\sqrt{36}=6\), so the candidate is valid.

The radical is already isolated; square both complete sides, solve, then substitute back.

Square both sides to get \(4x+9=49\), so \(x=10\). The original check gives \(\sqrt{49}=7\), so the candidate is valid.

The radical is already isolated; square both complete sides, solve, then substitute back.

Square both sides to get \(6x+1=25\), so \(x=4\). The original check gives \(\sqrt{25}=5\), so the candidate is valid.

Move the outside term before squaring so the entire radical is alone.

First isolate the radical: \(\sqrt{2x+1}=5\). Squaring and solving gives \(x=12\); substitution in the original equation verifies it.

Move the outside term before squaring so the entire radical is alone.

First isolate the radical: \(\sqrt{3x+4}=7\). Squaring and solving gives \(x=15\); substitution in the original equation verifies it.

Move the outside term before squaring so the entire radical is alone.

First isolate the radical: \(\sqrt{4x-3}=5\). Squaring and solving gives \(x=7\); substitution in the original equation verifies it.

Move the outside term before squaring so the entire radical is alone.

First isolate the radical: \(\sqrt{5x+6}=6\). Squaring and solving gives \(x=6\); substitution in the original equation verifies it.

Move the outside term before squaring so the entire radical is alone.

First isolate the radical: \(\sqrt{2x+7}=7\). Squaring and solving gives \(x=21\); substitution in the original equation verifies it.

Treat roots of the squared equation as candidates and test every one in the original equation.

Check the original equation: for \(x=3\), both sides equal \(3\), so it is valid. For \(x=-2\), the principal square root is nonnegative but the right side is \(-2\), so the equality fails and that candidate is extraneous.

Treat roots of the squared equation as candidates and test every one in the original equation.

Check the original equation: for \(x=2\), both sides equal \(3\), so it is valid. For \(x=-3\), the principal square root is nonnegative but the right side is \(-2\), so the equality fails and that candidate is extraneous.

Treat roots of the squared equation as candidates and test every one in the original equation.

Check the original equation: for \(x=4\), both sides equal \(3\), so it is valid. For \(x=-1\), the principal square root is nonnegative but the right side is \(-2\), so the equality fails and that candidate is extraneous.

Treat roots of the squared equation as candidates and test every one in the original equation.

Check the original equation: for \(x=1\), both sides equal \(3\), so it is valid. For \(x=-4\), the principal square root is nonnegative but the right side is \(-2\), so the equality fails and that candidate is extraneous.

Treat roots of the squared equation as candidates and test every one in the original equation.

Check the original equation: for \(x=6\), both sides equal \(4\), so it is valid. For \(x=-1\), the principal square root is nonnegative but the right side is \(-3\), so the equality fails and that candidate is extraneous.

Reject a candidate only after checking it in the original radical equation.

Substitution shows \(x=3\) satisfies the original equation, while \(x=-2\) makes its nonnegative left side unequal to the right side. Therefore \(-2\) is extraneous.

Reject a candidate only after checking it in the original radical equation.

Substitution shows \(x=2\) satisfies the original equation, while \(x=-3\) makes its nonnegative left side unequal to the right side. Therefore \(-3\) is extraneous.

Reject a candidate only after checking it in the original radical equation.

Substitution shows \(x=4\) satisfies the original equation, while \(x=-1\) makes its nonnegative left side unequal to the right side. Therefore \(-1\) is extraneous.

Reject a candidate only after checking it in the original radical equation.

Substitution shows \(x=1\) satisfies the original equation, while \(x=-4\) makes its nonnegative left side unequal to the right side. Therefore \(-4\) is extraneous.

Reject a candidate only after checking it in the original radical equation.

Substitution shows \(x=6\) satisfies the original equation, while \(x=-1\) makes its nonnegative left side unequal to the right side. Therefore \(-1\) is extraneous.

Set a square-root radicand greater than or equal to zero and reverse the inequality when dividing by a negative coefficient.

Require the radicand to be nonnegative: \(2x-6\ge0\). Solving that inequality gives \(x\ge3\).

Set a square-root radicand greater than or equal to zero and reverse the inequality when dividing by a negative coefficient.

Require the radicand to be nonnegative: \(-3x+12\ge0\). Solving that inequality gives \(x\le4\).

Set a square-root radicand greater than or equal to zero and reverse the inequality when dividing by a negative coefficient.

Require the radicand to be nonnegative: \(5x+10\ge0\). Solving that inequality gives \(x\ge-2\).

Set a square-root radicand greater than or equal to zero and reverse the inequality when dividing by a negative coefficient.

Require the radicand to be nonnegative: \(-4x-8\ge0\). Solving that inequality gives \(x\le-2\).

Set a square-root radicand greater than or equal to zero and reverse the inequality when dividing by a negative coefficient.

Require the radicand to be nonnegative: \(7x-1\ge0\). Solving that inequality gives \(x\ge\frac17\).

Cube both sides of an isolated cube-root equation; odd powers are one-to-one over the reals.

Cubing both sides gives \(2x+5=27\), so \(x=11\). Substitution gives cube root \(3\), confirming the value.

Cube both sides of an isolated cube-root equation; odd powers are one-to-one over the reals.

Cubing both sides gives \(3x-1=-8\), so \(x=-2.3333333333333335\). Substitution gives cube root \(-2\), confirming the value.

Cube both sides of an isolated cube-root equation; odd powers are one-to-one over the reals.

Cubing both sides gives \(4x+7=1\), so \(x=-1.5\). Substitution gives cube root \(1\), confirming the value.

Cube both sides of an isolated cube-root equation; odd powers are one-to-one over the reals.

Cubing both sides gives \(5x-10=-27\), so \(x=-3.4\). Substitution gives cube root \(-3\), confirming the value.

Cube both sides of an isolated cube-root equation; odd powers are one-to-one over the reals.

Cubing both sides gives \(6x+2=8\), so \(x=1\). Substitution gives cube root \(2\), confirming the value.

Do not reject a negative candidate automatically; check whether the original right side and radical value agree.

Squaring and factoring yields candidates \(0\) and \(1\). Direct substitution into the original equation makes both sides equal for each candidate, so both remain in the solution set.

Do not reject a negative candidate automatically; check whether the original right side and radical value agree.

Squaring and factoring yields candidates \(0\) and \(2\). Direct substitution into the original equation makes both sides equal for each candidate, so both remain in the solution set.

Do not reject a negative candidate automatically; check whether the original right side and radical value agree.

Squaring and factoring yields candidates \(0\) and \(4\). Direct substitution into the original equation makes both sides equal for each candidate, so both remain in the solution set.

Do not reject a negative candidate automatically; check whether the original right side and radical value agree.

Squaring and factoring yields candidates \(0\) and \(1\). Direct substitution into the original equation makes both sides equal for each candidate, so both remain in the solution set.

Do not reject a negative candidate automatically; check whether the original right side and radical value agree.

Squaring and factoring yields candidates \(-1\) and \(0\). Direct substitution into the original equation makes both sides equal for each candidate, so both remain in the solution set.

Expand the entire binomial square; never square its terms separately and omit the cross term.

Use \((a+b)^2=a^2+2ab+b^2\), including the cross term. Therefore \((x+2)^2=x^2+4x+4\).

Expand the entire binomial square; never square its terms separately and omit the cross term.

Use \((a+b)^2=a^2+2ab+b^2\), including the cross term. Therefore \((x-3)^2=x^2-6x+9\).

Expand the entire binomial square; never square its terms separately and omit the cross term.

Use \((a+b)^2=a^2+2ab+b^2\), including the cross term. Therefore \((2x+1)^2=4x^2+4x+1\).

Expand the entire binomial square; never square its terms separately and omit the cross term.

Use \((a+b)^2=a^2+2ab+b^2\), including the cross term. Therefore \((3-x)^2=x^2-6x+9\).

Expand the entire binomial square; never square its terms separately and omit the cross term.

Use \((a+b)^2=a^2+2ab+b^2\), including the cross term. Therefore \((x+5)^2=x^2+10x+25\).

Check the sign and exact value of the original nonradical side for every candidate.

For \(x=2\), both original sides equal \(1\). For \(x=5\), the radical is nonnegative while \(3-x=-2\), so the equality fails. Thus \(5\) is extraneous.

Check the sign and exact value of the original nonradical side for every candidate.

For \(x=3\), both original sides equal \(1\). For \(x=6\), the radical is nonnegative while \(4-x=-2\), so the equality fails. Thus \(6\) is extraneous.

Check the sign and exact value of the original nonradical side for every candidate.

For \(x=4\), both original sides equal \(1\). For \(x=7\), the radical is nonnegative while \(5-x=-2\), so the equality fails. Thus \(7\) is extraneous.

Check the sign and exact value of the original nonradical side for every candidate.

For \(x=5\), both original sides equal \(1\). For \(x=8\), the radical is nonnegative while \(6-x=-2\), so the equality fails. Thus \(8\) is extraneous.

Check the sign and exact value of the original nonradical side for every candidate.

For \(x=6\), both original sides equal \(1\). For \(x=9\), the radical is nonnegative while \(7-x=-2\), so the equality fails. Thus \(9\) is extraneous.

Identify the exact step where equivalence may be lost or algebra was applied incompletely.

A safe radical-equation solution isolates the radical, raises both complete sides, solves, and checks every candidate in the original equation. The stated correction follows that workflow.

Identify the exact step where equivalence may be lost or algebra was applied incompletely.

A safe radical-equation solution isolates the radical, raises both complete sides, solves, and checks every candidate in the original equation. The stated correction follows that workflow.

Identify the exact step where equivalence may be lost or algebra was applied incompletely.

A safe radical-equation solution isolates the radical, raises both complete sides, solves, and checks every candidate in the original equation. The stated correction follows that workflow.

Identify the exact step where equivalence may be lost or algebra was applied incompletely.

A safe radical-equation solution isolates the radical, raises both complete sides, solves, and checks every candidate in the original equation. The stated correction follows that workflow.

Identify the exact step where equivalence may be lost or algebra was applied incompletely.

A safe radical-equation solution isolates the radical, raises both complete sides, solves, and checks every candidate in the original equation. The stated correction follows that workflow.