Solve Where x Is a Real Number: Full Solution to sqrt(3x + 4) = x

Problem. Solve $\sqrt{3x+4}=x$ where x is a real number.

One radical, an $x$ on the other side, and the real-number clause at the end. When a test says solve where x is a real number, it is announcing: some numbers you are about to produce will not survive. To solve where x is a real number: isolate the radical, square, solve, then check every candidate in the original equation. The strip below previews the finish: solve where x is a real number eliminates one candidate on the spot.

Answer. $x=4$.

Squaring also produces $x=-1$, but $\sqrt{3(-1)+4}=\sqrt{1}=1$ while the right side is $-1$. A principal square root is never negative, so $-1$ is thrown out. That single line is the whole point: when you solve where x is a real number, the answer list shrinks from two names to one. This is the solve where x is a real number answer, settled up front.

Durell's 1911 School Algebra states the machine: transpose so a single radical stands alone, raise both sides to the power of the index, repeat if radicals remain. Add the key move - check in the original equation. Run all four.

Step 1, isolate. The radical $\sqrt{3x+4}$ already sits alone.

Step 2, square both sides. $3x+4=x^2$.

Step 3, solve the quadratic. $x^2-3x-4=0$, so $(x-4)(x+1)=0$. Candidates: $x=4$ and $x=-1$.

Step 4, check each candidate - the solve where x is a real number step. For $x=4$: $\sqrt{16}=4$ against 4 - true. For $x=-1$: $\sqrt{1}=1$ against $-1$ - false, reject.

Why did the fake appear? Squaring erases signs: $\sqrt{4}=2$, never $-2$. Squaring also catches the shadow equation $\sqrt{3x+4}=-x$, which $-1$ solves but you never wrote. Durell saw it in 1911 - flip the radical's sign in $5+\sqrt{x}=2$ and $x=9$ becomes provable. When you solve where x is a real number, the check defines the answer, not hygiene.

Problem. Solve $\sqrt{x^2+7}-1=x$ where x is a real number. (Durell, School Algebra, 1911, Art. 215.)

Step 1, isolate the radical. $\sqrt{x^2+7}=x+1$.

Step 2, square. $x^2+7=x^2+2x+1$.

Step 3, solve. The $x^2$ terms cancel: $7=2x+1$, so $x=3$.

Step 4, check. Left: $\sqrt{16}=4$. Right: $3+1=4$. True.

One candidate, one survivor. Even with no impostor, solve where x is a real number still ends with the check - Durell's note: use only the principal value of the radical.

Problem. Solve $\sqrt{x+3}+\sqrt{x}=5$ where x is a real number. (Durell, Art. 215, Example 2.)

Step 1, isolate one radical. $\sqrt{x+3}=5-\sqrt{x}$.

Step 2, square. $x+3=25-10\sqrt{x}+x$, so $\sqrt{x}=\frac{11}{5}$.

Step 3, square again. $x=\frac{121}{25}$.

Step 4, check. $\sqrt{\frac{196}{25}}=\frac{14}{5}$ and $\sqrt{\frac{121}{25}}=\frac{11}{5}$; the sum is $\frac{25}{5}=5$. True.

Two squarings, two chances for impostors - both pass here. And $\frac{121}{25}=4.84$ shows that when you solve where x is a real number, roots need not be whole; they must only survive the check.

1. Skipping the check. Squaring created $-1$ out of thin air above, and only the check exposed it. Every time you solve where x is a real number, the check is the last step.

2. Believing $\sqrt{a+b}=\sqrt{a}+\sqrt{b}$. Durell's own trap: $\sqrt{9+16}=\sqrt{25}=5$, not $3+4=7$. A root of a sum is not the sum of the roots.

3. Forgetting the principal root. $\sqrt{4}=2$, never $-2$. To solve where x is a real number, a candidate that makes the radical side negative is automatically extraneous - reject it on sight.

4. Announcing two answers. After $(x-4)(x+1)=0$, the slip is listing both values. The whole job of solve where x is a real number is shrinking that list to the survivors.

5. Squaring before isolating. From $\sqrt{x+3}+\sqrt{x}=5$, squaring in place buries the cross term $2\sqrt{(x+3)x}$. Isolate one radical first - Durell's rule.