Solving Radical Equations - Algebra 3

1. The equation $\sqrt{2x+3}=x+a$ has exactly one real solution, and that solution is $x=-1$. What is the value of $a$?
   Answer: $2$ Substitute $x=-1$: $\sqrt{1}=-1+a$, so $a=2$. With $a=2$, squaring gives $2x+3=x^2+4x+4$, or $(x+1)^2=0$, confirming exactly one solution. $-2$ comes from a sign error in solving $1=a-1$; $1$ and $4$ do not make $x=-1$ a solution.
2. What is the solution set of $\sqrt{3x+1}-\sqrt{x+1}=2$?
   Answer: $\{8\}$ Domain $x\ge -\frac{1}{3}$. Isolate $\sqrt{3x+1}=2+\sqrt{x+1}$. Squaring gives $3x+1=x+5+4\sqrt{x+1}$, so $x-2=2\sqrt{x+1}$. Squaring again gives $x^2-8x=0$, so $x=0$ or $x=8$. Checking: $x=0$ gives $1-1=0$, not 2, so it is extraneous; $x=8$ gives $5-3=2$. The solution set is $\{8\}$. $\{0\}$ keeps the extraneous root, $\{0,8\}$ fails to check, and $\varnothing$ incorrectly rejects the valid root.
3. Solve $\sqrt{3x+4}-\sqrt{x+2}=\sqrt{x}$.
   Answer: $\left\{\frac{2}{3}\right\}$ Domain $x\ge0$. Isolate $\sqrt{3x+4}=\sqrt{x+2}+\sqrt{x}$. Squaring gives $3x+4=2x+2+2\sqrt{x(x+2)}$, so $x+2=2\sqrt{x^2+2x}$. Squaring again gives $3x^2+4x-4=0$, with candidates $x=\frac23$ and $x=-2$. $x=-2$ is outside the domain; $x=\frac23$ checks. $\left\{-2,\frac23\right\}$ includes the domain violation, $\{-2\}$ keeps only the invalid candidate, and $\varnothing$ rejects the valid solution.
4. How many real solutions does the equation $\sqrt{2x+3}-\sqrt{x-1}=1$ have?
   Answer: 0 Domain: $x\ge 1$. Isolate: $\sqrt{2x+3}=1+\sqrt{x-1}$. Squaring gives $2x+3=x+2\sqrt{x-1}$, so $x+3=2\sqrt{x-1}$. Squaring again gives $x^2+2x+13=0$, whose discriminant is $4-52=-48<0$. Thus there are no real candidates to check, so the equation has 0 real solutions. The other counts would require real roots of the final quadratic, but its discriminant is negative.
5. For which value of $k$ does $\sqrt{x+2}=x+k$ have no real solution?
   Answer: 3 Squaring gives $x^2+(2k-1)x+(k^2-2)=0$. Its discriminant is $(2k-1)^2-4(k^2-2)=9-4k$. For $k=3$, the discriminant is $-3<0$, so the squared equation has no real roots. For $k=0,1,2$, the discriminant is nonnegative and at least one resulting candidate checks in the original equation.
6. Solve $(3x+4)^{1/2}=x-2$. What is the real solution set?
   Answer: $\{7\}$ When you see a radical equation like $(3x+4)^{1/2}=x-2$, remember that the radical sign means the principal (nonnegative) square root. Squaring both sides is a valid step, but it can create extraneous solutions, so every candidate must be checked in the original equation. First, square both sides: $(3x+4)=(x-2)^2$. Expanding gives $3x+4=x^2-4x+4$, so $x^2-7x=0$, meaning $x=0$ or $x=7$. Now check each candidate in the original equation. For $x=0$, the left side is $(3(0)+4)^{1/2}=\sqrt{4}=2$, but the right side is $0-2=-2$. Since $2 \ne -2$, $x=0$ is extraneous. For $x=7$, the left side is $(21+4)^{1/2}=\sqrt{25}=5$, and the right side is $7-2=5$. So the real solution set is $\{7\}$. The choice $\{0\}$ is wrong because it keeps only the extraneous value of $0$, likely from forgetting that $\sqrt{4}$ cannot equal $-2$. The choice $\{0,7\}$ is wrong because it includes $0$ without checking; this is the classic trap of solving by squaring and stopping too soon. The empty set is wrong because it misses the valid solution $7$; perhaps the student noticed $0$ failed and assumed no solution instead of checking the other candidate. Always isolate the radical, square, solve, and then test every candidate in the original equation. If a candidate fails, discard it — but only after checking all possibilities.
7. Which equation has $x=2$ as an extraneous solution?
   Answer: $\sqrt{x+7}=x-5$ For $\sqrt{x+7}=x-5$, squaring gives $(x-2)(x-9)=0$. At $x=2$, the left side is $3$ but the right side is $-3$, so $x=2$ is extraneous. In the other equations, $x=2$ either checks or is not a candidate produced by squaring.
8. Solve $\sqrt{2x+3}+\sqrt{x+1}=1$.
   Answer: $\{-1\}$ Domain $x\ge -1$. At $x=-1$, the left side is $1+0=1$. For $x>-1$, $\sqrt{2x+3}>1$, so the left side is greater than 1. Algebraically, isolating and squaring leads to $x+1=-2\sqrt{x+1}$, which forces $x=-1$; the candidate $x=3$ from squaring again is extraneous. $\{3\}$ keeps the extraneous candidate, $\{-1,3\}$ fails the sign check, and $\varnothing$ misses the boundary solution.
9. What is the solution set of $\sqrt{x+4}=\sqrt{x}-2$?
   Answer: $\varnothing$ Whenever you solve an equation involving square roots, the key is to remember that squaring both sides can create extraneous solutions. So after solving, you must check every candidate in the original equation. Here, set $y=\sqrt{x}$, so $y\ge 0$. The equation becomes $$\sqrt{y^2+4}=y-2.$$ The right side $y-2$ must be nonnegative, so $y\ge 2$. Squaring both sides gives $$y^2+4=y^2-4y+4 \quad\Rightarrow\quad 0=-4y \quad\Rightarrow\quad y=0.$$ But $y=0$ contradicts $y\ge 2$, so there is no solution. Checking the proposed answers confirms this: $\{0\}$ comes from the extraneous result $y=0$, but substituting $x=0$ gives $\sqrt{4}=2$ on the left and $\sqrt{0}-2=-2$ on the right. $\{4\}$ fails too, since $\sqrt{8}\neq 0$. And $\{0,4\}$ just combines two non-solutions. Therefore the solution set is $\varnothing$. Study tip: after squaring a radical equation, always plug your answers back into the original equation, or check the required sign condition before squaring. That will quickly eliminate extraneous roots like $x=0$ here.
10. What is the solution set of $\sqrt{x^2+3x+5}=x+1$?
    Answer: $\varnothing$ Whenever you solve a radical equation, remember that squaring both sides is not reversible: it can create extraneous solutions. Also, the square root symbol means the nonnegative root, so the right-hand side $x+1$ must be nonnegative too; that is, $x \ge -1$. Square both sides anyway: $$\sqrt{x^2+3x+5}=x+1$$ $$x^2+3x+5=(x+1)^2=x^2+2x+1$$ \[ 3x+5=2x+1 \implies x=-4. But \(x=-4\) violates the necessary condition \(x\ge -1\). Check the original too: the left side becomes \(\sqrt{9}=3\), while the right side becomes \(-4+1=-3\). They do not match, so \(-4\) is extraneous. Therefore no real number satisfies the equation; the solution set is empty. The choice \(\{-4\}\) is the classic trap: it comes from squaring and solving but forgetting to check whether the candidate actually works. The choice \(\{-4,4\}\) is also impossible because squaring produced exactly one candidate, so \(4\) cannot be added as a solution. And the choice \(\{4\}\) fails direct substitution:when \(x=4\), left side is \(\sqrt{33}\), not \(5\). These errors usually come from sign mistakes or from treating extraneous roots as valid. Your takeaway: in any radical equation, after squaring, always test every candidate in the original equation. Even better, before squaring, impose the condition that the non-radical side must be nonnegative;this catches most extraneous solutions immediately.
11. What is the solution set of $\sqrt{x+3}-\sqrt{2x-1}=-1$?
    Answer: $\{13\}$ When you see a radical equation, the core danger is extraneous solutions: squaring both sides can create answers that do not actually satisfy the original equation. So your job is not just to solve, but to verify every candidate in the original equation. Start by isolating one radical: $$\sqrt{x+3} = \sqrt{2x-1}-1$$ Wait, better to isolate the more convenient radical. From $$\sqrt{x+3}-\sqrt{2x-1}=-1,$$ move one radical: $$\sqrt{x+3}+1=\sqrt{2x-1}.$$ Square both sides: $$x+3+2\sqrt{x+3}+1=2x-1,$$ so $$2\sqrt{x+3}=x-5.$$ Notice this already requires $x\ge 5$, because the left side is nonnegative. Square again: $$4(x+3)=(x-5)^2$$ $$4x+12=x^2-10x+25$$ $$x^2-14x+13=0$$ $$(x-1)(x-13)=0.$$ So candidates are $1$ and $13$. But $x=1$ fails the condition $x\ge5$, and checking the original gives $\sqrt4-\sqrt1=1$, not $-1$. So $1$ is extraneous. Checking $x=13$: $$\sqrt{16}-\sqrt{25}=4-5=-1.$$ Thus the solution set is $\{13\}$. The choice $\{1\}$ comes from forgetting to verify. The choice $\{1,13\}$ includes the extraneous root created by squaring. The choice $\varnothing$ incorrectly assumes a negative result is impossible, but one radical can be smaller than the other by 1. Study tip: after solving any radical equation, always substitute your answers back into the original equation. Squaring can introduce false solutions, and checking is the only reliable filter.
12. A student correctly squares $\sqrt{x+6}=x+4$ and obtains the candidates $x=-5$ and $x=-2$. Which value or values are extraneous?
    Answer: Only $x=-5$ Whenever you square both sides of an equation, you can create extraneous solutions: squaring is not reversible, because $a=b$ implies $a^2=b^2$, but $a^2=b^2$ does not imply $a=b$. So after solving the squared equation, you must plug each candidate back into the original radical equation. For $x=-5$: $$\sqrt{-5+6}=\sqrt{1}=1,$$ but the right side is $$-5+4=-1.$$ Since $1 \ne -1$, $x=-5$ fails and is extraneous. For $x=-2$: $$\sqrt{-2+6}=\sqrt{4}=2,$$ and the right side is $$-2+4=2.$$ This matches, so $x=-2$ is valid. Thus "Only $x=-5$" is correct. "Only $x=-2$" mistakes the valid solution for the extraneous one. "Both $x=-5$ and $x=-2$" assumes squaring always invalidates both candidates, but only the one that fails the original check is extraneous. "Neither $x=-5$ nor $x=-2$" misses that squaring can introduce an extra solution, and $x=-5$ is exactly such a false solution because it makes the original equation have $1$ on the left and $-1$ on the right. On any radical equation, make it a habit to substitute your candidates into the original equation. If a candidate satisfies the original equation, keep it; if not, discard it as extraneous.
13. How many real solutions does $(x^2+3x+4)^{1/2}=x+1$ have?
    Answer: 0 Squaring gives $x^2+3x+4=x^2+2x+1$, so $x=-3$. But in the original equation the right side is $-2$ at $x=-3$, while the radical is nonnegative; indeed $\sqrt{4}=2$, so $x=-3$ is extraneous. With no other candidates, there are 0 real solutions. 1 ignores the sign check; 2 or 3 would require additional candidates that squaring does not produce.
14. What is the real solution of $\sqrt{x+5}+\sqrt{x-2}=4$?
    Answer: $\left\{\frac{209}{64}\right\}$ When you see a radical equation like $\sqrt{x+5}+\sqrt{x-2}=4$, your first move should be to isolate a radical, then square both sides. But because squaring can introduce extraneous solutions, always check your final answer in the original equation. Start by isolating: $\sqrt{x+5}=4-\sqrt{x-2}$. Squaring gives $x+5=16-8\sqrt{x-2}+x-2$. Simplify: $x+5=x+14-8\sqrt{x-2}$, so $-9=-8\sqrt{x-2}$, meaning $\sqrt{x-2}=\frac{9}{8}$. Squaring again yields $x-2=\frac{81}{64}$, so $x=\frac{209}{64}$. Checking in the original equation confirms it works: $\sqrt{\frac{209}{64}+5}+\sqrt{\frac{209}{64}-2}=\frac{23}{8}+\frac{9}{8}=4$. The choice $\frac{81}{64}$ is a trap: it is the value of $x-2$, not the value of $x$. You must add the 2 back to solve for $x$. Similarly, $\frac{9}{8}$ is the value of $\sqrt{x-2}$ after the first squaring step, not the final solution for $x$. Finally, $\varnothing$ is incorrect because the solution $\frac{209}{64}$ is real and satisfies the original equation; there is no extraneous root here. Your study tip: whenever you square a radical equation, finish by substituting your candidates back into the original equation. That check is what separates valid solutions from extraneous ones.
15. What are the real solutions of $x-4\sqrt{x}+3=0$?
    Answer: $\{1,9\}$ Let $y=\sqrt{x}\ge0$. Then $y^2-4y+3=0$, so $y=1$ or $y=3$. Since $x=y^2$, $x=1$ or $x=9$. Both satisfy the original equation. $\{1,3\}$ confuses $y$-values with $x$-values; $\{3,9\}$ mixes one $y$-value with a solution; $\{9\}$ loses the $y=1$ case.
16. Solve $\sqrt{2x+3}-\sqrt{x+1}=1$.
    Answer: $\{-1,3\}$ Domain $x\ge -1$. Isolate and square: $2x+3=x+2+2\sqrt{x+1}$, so $x+1=2\sqrt{x+1}$. Let $y=\sqrt{x+1}\ge0$; then $y^2=2y$, so $y=0$ or $y=2$. Hence $x=-1$ or $x=3$. Both check in the original equation. $\{3\}$ comes from dividing by $y$ and losing $y=0$; $\{-1\}$ drops the valid root; $\varnothing$ ignores both checks.
17. Which equation has two real solutions?
    Answer: $\sqrt{x+2}=x+2$ For $\sqrt{x+2}=x+2$, squaring gives $(x+2)^2=x+2$, so $(x+2)(x+1)=0$, yielding $x=-2$ and $x=-1$; both check. $\sqrt{x+2}=x$ has only $x=2$ after rejecting $x=-1$; $\sqrt{x+2}=x+1$ has only $\frac{-1+\sqrt5}{2}$ after rejecting the negative candidate; $\sqrt{x+2}=x+3$ has a negative discriminant and no solution.
18. Solve $\sqrt{2x+1}=-x$. What is the real solution set?
    Answer: $\{1-\sqrt{2}\}$ Whenever you solve a radical equation like $\sqrt{2x+1}=-x$, remember that the square root symbol means the nonnegative root. So the right-hand side, $-x$, must also be nonnegative, which means $x \le 0$. Also, the radicand requires $2x+1 \ge 0$, so $x \ge -\frac12$. Any valid solution must lie in $[-\tfrac12, 0]$. Squaring both sides gives $2x+1 = x^2$, or $x^2 - 2x - 1 = 0$. By the quadratic formula, $x = 1 \pm \sqrt{2}$. Now test these against the original equation. The value $1+\sqrt{2} \approx 2.414$ is positive, so it violates the requirement that $-x$ be nonnegative; it is an extraneous root introduced by squaring. The value $1-\sqrt{2} \approx -0.414$ is in the valid interval, and checking: $\sqrt{2(1-\sqrt2)+1} = -(1-\sqrt2)$ works. So the real solution set is $\{1-\sqrt{2}\}$. The choice listing $1+\sqrt{2}$ alone misses the domain check. The choice with both values includes the extraneous positive root. The empty set is the trap if you assume $-x$ is always negative; actually, when $x$ is negative, $-x$ is positive, so a radical can equal it. Your study tip: after squaring a radical equation, always check every candidate in the original equation, and use sign restrictions before solving to filter out extraneous answers quickly.
19. Solve $\sqrt{x^2+4x+4}=x+6$. What is the real solution set?
    Answer: $\{-4\}$ When you see a square root containing a perfect-square quadratic, your first thought should be: $\sqrt{a^2}=|a|$. Here, $x^2+4x+4=(x+2)^2$, so the equation becomes $$\sqrt{(x+2)^2}=x+6 \quad \Rightarrow \quad |x+2|=x+6.$$ Now split into cases. If $x+2\ge 0$, then $x+2=x+6$, which gives $2=6$, impossible. If $x+2<0$, then $-(x+2)=x+6$, so $$-x-2=x+6 \quad \Rightarrow \quad -2x=8 \quad \Rightarrow \quad x=-4.$$ Check $x=-4$: the left side is $\sqrt{16-16+4}=2$, and the right side is $-4+6=2$. So $\{-4\}$ is the real solution set. The choice $\{-2\}$ fails because plugging in $-2$ gives $0$ on the left but $4$ on the right; it only makes the radicand zero, not the whole equation true. The choice $\{-4,-2\}$ includes the valid $-4$ but repeats that same extraneous $-2$. The choice $\varnothing$ comes from mistakenly simplifying $\sqrt{(x+2)^2}$ to $x+2$ instead of $|x+2|$; that leads to $x+2=x+6$, which has no solution, but the absolute-value equation does have one. Always rewrite $\sqrt{\text{perfect square}}$ as an absolute value, and check every candidate in the original equation before finalizing your solution set.
20. Solve $\sqrt{x+3}=x+3$.
    Answer: $\{-3,-2\}$ Let $y=x+3$. Then $\sqrt{y}=y$. Since $y\ge0$, squaring gives $y=y^2$, so $y=0$ or $y=1$. Thus $x=-3$ or $x=-2$, and both check. $\{-3\}$ and $\{-2\}$ each lose one solution; $\varnothing$ ignores that $\sqrt0=0$ and $\sqrt1=1$.

Start this quiz