Exam practice

Squaring: 2x = 36, x = 18. Verify: $\sqrt{36} = 6$ ✓. x = 18.

Squaring: 3x − 5 = 1, 3x = 6, x = 2. Verify: $\sqrt{6 - 5} = \sqrt{1} = 1$ ✓. x = 2.

Separate: $\sqrt{3x + 4} = 5$. Squaring: 3x + 4 = 25, 3x = 21, x = 7. Verify: $6 - \sqrt{25} = 6 - 5 = 1$ ✓. x = 7.

Cubing: 5x − 3 = 27, 5x = 30, x = 6. Verify: $\sqrt[3]{30 - 3} = \sqrt[3]{27} = 3$ ✓. x = 6.

Separate: $\sqrt{3x + 6} = 3\sqrt{x - 4}$. Squaring:
$$\begin{aligned} 3x + 6 &= 9(x - 4) \\ 3x + 6 &= 9x - 36 \\ 6x &= 42 \\ x &= 7 \end{aligned}$$
Verify: $\sqrt{27} - 3\sqrt{3} = 3\sqrt{3} - 3\sqrt{3} = 0$ ✓. x = 7.

Squaring: $11x^2 + 45 = 16x^2$, so $5x^2 = 45$, $x^2 = 9$, x = ±3.
Verify x = 3: $\sqrt{99 + 45} = \sqrt{144} = 12 = 4 \times 3$ ✓.
Verify x = −3: $\sqrt{144} = 12$, but 4 × (−3) = −12 ✗.
x = 3 (not −3).

Separate: $\sqrt{x + 7} = 7 - \sqrt{x}$. Squaring:
$$\begin{aligned} x + 7 &= 49 - 14\sqrt{x} + x \\ 14\sqrt{x} &= 42 \\ \sqrt{x} &= 3 \\ x &= 9 \end{aligned}$$
Verify: $\sqrt{16} + \sqrt{9} = 4 + 3 = 7$ ✓. x = 9.

Squaring:
$$\begin{aligned} x + 5 &= 25 - 10\sqrt{x} + x \\ 10\sqrt{x} &= 20 \\ \sqrt{x} &= 2 \\ x &= 4 \end{aligned}$$
Verify: $\sqrt{9} = 3$ and $5 - \sqrt{4} = 3$ ✓. x = 4.

After the first squaring you get $x - 2 = 2\sqrt{x - 2}$. Don’t divide by $\sqrt{x - 2}$: it might be 0.

Separate: $\sqrt{3x - 2} = 2 + \sqrt{x - 2}$. Squaring:
$$\begin{aligned} 3x - 2 &= 4 + 4\sqrt{x - 2} + x - 2 \\ 2x - 4 &= 4\sqrt{x - 2} \\ x - 2 &= 2\sqrt{x - 2} \end{aligned}$$
Squaring again:
$$\begin{aligned} (x - 2)^2 &= 4(x - 2) \\ (x - 2)(x - 2 - 4) &= 0 \\ (x - 2)(x - 6) &= 0 \end{aligned}$$
so x = 2 or 6.
Verify x = 2: $\sqrt{4} - \sqrt{0} = 2$ ✓. Verify x = 6: $\sqrt{16} - \sqrt{4} = 4 - 2 = 2$ ✓.
x = 2 or 6.

This is Example 1c. $\sqrt{3x + 10} = x$; squaring, $x^2 - 3x - 10 = 0$, (x − 5)(x + 2) = 0, x = 5 or −2.
Verify x = 5: $\sqrt{25} - 5 = 0$ ✓. Verify x = −2: $\sqrt{4} + 2 = 4 \ne 0$ ✗.
x = 5 (not −2).

Squaring:
$$\begin{aligned} 3x - 5 &= x^2 - 2x + 1 \\ x^2 - 5x + 6 &= 0 \\ (x - 2)(x - 3) &= 0 \end{aligned}$$
so x = 2 or 3.
Verify x = 2: $\sqrt{1} = 1 = 2 - 1$ ✓. Verify x = 3: $\sqrt{4} = 2 = 3 - 1$ ✓.
x = 2 or 3: both true roots.

Squaring:
$$\begin{aligned} 2x^2 - 7 &= x^2 + 6x + 9 \\ x^2 - 6x - 16 &= 0 \\ (x - 8)(x + 2) &= 0 \end{aligned}$$
so x = 8 or −2.
Verify x = 8: $\sqrt{128 - 7} = \sqrt{121} = 11 = 8 + 3$ ✓. Verify x = −2: $\sqrt{8 - 7} = 1 = -2 + 3$ ✓.
x = 8 or −2: here the negative answer is a true root, since the right side x + 3 is still positive.

Multiply both sides by $\sqrt{x}$ first.

Multiplying both sides by $\sqrt{x}$ (x > 0):
$$\sqrt{x(x - 1)} + x = 2 \quad\Rightarrow\quad \sqrt{x^2 - x} = 2 - x$$
Squaring:
$$\begin{aligned} x^2 - x &= 4 - 4x + x^2 \\ 3x &= 4 \\ x &= 4/3 \end{aligned}$$
Verify: $\sqrt{1/3} + \sqrt{4/3} = \dfrac{1}{\sqrt{3}} + \dfrac{2}{\sqrt{3}} = \dfrac{3}{\sqrt{3}} = \sqrt{3}$, and $\dfrac{2}{\sqrt{4/3}} = \dfrac{2\sqrt{3}}{2} = \sqrt{3}$ ✓. x = 4/3.

$x - 1 = (\sqrt{x} + 1)(\sqrt{x} - 1)$, as in Example 2, but this time it is $\sqrt{x} + 1$ that cancels.

The left side is $\dfrac{(\sqrt{x} + 1)(\sqrt{x} - 1)}{\sqrt{x} + 1} = \sqrt{x} - 1$. So
$$\begin{aligned} \sqrt{x} - 1 &= 3 + \frac{\sqrt{x} + 1}{2} \\ 2\sqrt{x} - 2 &= 6 + \sqrt{x} + 1 \\ \sqrt{x} &= 9 \\ x &= 81 \end{aligned}$$
Verify: $\dfrac{80}{9 + 1} = 8$, and $3 + \dfrac{9 + 1}{2} = 3 + 5 = 8$ ✓. x = 81.
(With $\sqrt{x} - 1$ at the bottom instead, this is Example 2, whose answer is 25.)

This is Example 3. Squaring twice gives $3x^2 - 16x + 16 = 0$, (3x − 4)(x − 4) = 0, so x = 4 or 4/3.
Verify x = 4: $\sqrt{9} - \sqrt{4} = 1$ ✓. Verify x = 4/3: $\sqrt{25/9} - \sqrt{4/9} = 5/3 - 2/3 = 1$ ✓.
x = 4 or 4/3.

Cross-multiply, or use componendo and dividendo: if a/b = c/d then (a + b)/(a − b) = (c + d)/(c − d).

Cross-multiplying:
$$\begin{aligned} \sqrt{x + 4} + \sqrt{x - 4} &= 2\sqrt{x + 4} - 2\sqrt{x - 4} \\ 3\sqrt{x - 4} &= \sqrt{x + 4} \end{aligned}$$
Squaring:
$$\begin{aligned} 9(x - 4) &= x + 4 \\ 8x &= 40 \\ x &= 5 \end{aligned}$$
Verify: $\sqrt{9} = 3$, $\sqrt{1} = 1$, and $\dfrac{3 + 1}{3 - 1} = 2$ ✓. x = 5.

$\sqrt{4x + 8} = \sqrt{4(x + 2)} = 2\sqrt{x + 2}$.

Since $\sqrt{4x + 8} = 2\sqrt{x + 2}$:
$$\begin{aligned} \frac{1}{\sqrt{x + 2}} - \frac{1}{2\sqrt{x + 2}} &= 2 \\ \frac{1}{2\sqrt{x + 2}} &= 2 \\ \sqrt{x + 2} &= \frac{1}{4} \end{aligned}$$
Squaring: x + 2 = 1/16, so x = 1/16 − 2 = −31/16.
Verify: x + 2 = 1/16 and 4x + 8 = 1/4, so $\dfrac{1}{1/4} - \dfrac{1}{1/2} = 4 - 2 = 2$ ✓. x = −31/16.

Separate: $\sqrt{x^2 - 3x + 3} = 2 - \sqrt{x^2 - x + 1}$. Squaring:
$$\begin{aligned} x^2 - 3x + 3 &= 4 - 4\sqrt{x^2 - x + 1} + x^2 - x + 1 \\ 4\sqrt{x^2 - x + 1} &= 2x + 2 \\ 2\sqrt{x^2 - x + 1} &= x + 1 \end{aligned}$$
Squaring again:
$$\begin{aligned} 4x^2 - 4x + 4 &= x^2 + 2x + 1 \\ 3x^2 - 6x + 3 &= 0 \\ 3(x - 1)^2 &= 0 \end{aligned}$$
so x = 1.
Verify: $\sqrt{1 - 3 + 3} + \sqrt{1 - 1 + 1} = 1 + 1 = 2$ ✓. x = 1.

Separate: $\sqrt{x^2 - 2x - 4} = 1 + \sqrt{x^2 - 3x - 3}$. Squaring:
$$\begin{aligned} x^2 - 2x - 4 &= 1 + 2\sqrt{x^2 - 3x - 3} + x^2 - 3x - 3 \\ x - 2 &= 2\sqrt{x^2 - 3x - 3} \end{aligned}$$
Squaring again:
$$\begin{aligned} x^2 - 4x + 4 &= 4x^2 - 12x - 12 \\ 3x^2 - 8x - 16 &= 0 \\ (3x + 4)(x - 4) &= 0 \end{aligned}$$
so x = 4 or −4/3.
Verify x = 4: $\sqrt{16 - 8 - 4} - \sqrt{16 - 12 - 3} = 2 - 1 = 1$ ✓.
Verify x = −4/3: $\sqrt{16/9 + 8/3 - 4} - \sqrt{16/9 + 4 - 3} = \sqrt{4/9} - \sqrt{25/9} = 2/3 - 5/3 = -1 \ne 1$ ✗.
x = 4 (not −4/3). The sign shows it early: $x - 2 = 2\sqrt{\ldots}$ needs x − 2 ≥ 0, and −4/3 − 2 is negative.

Since $x - 1 = (\sqrt{x} + 1)(\sqrt{x} - 1)$, the left side is $\sqrt{x} - 1$:
$$\begin{aligned} \sqrt{x} - 1 &= 4 + \frac{\sqrt{x} - 1}{2} \\ 2\sqrt{x} - 2 &= 8 + \sqrt{x} - 1 \\ \sqrt{x} &= 9 \\ x &= 81 \end{aligned}$$
Verify: $\dfrac{81 - 1}{9 + 1} = 8$, and $4 + \dfrac{9 - 1}{2} = 4 + 4 = 8$ ✓. So x = 81. (Proved.)

The equation is undefined when x < 0: then $\sqrt{x}$ is not a real number. Its denominator $\sqrt{x} + 1$ is never 0, since $\sqrt{x} \ge 0$ makes it at least 1.

Your own five, each solved and checked. Five to compare with, one of each kind:

• $\sqrt{4x + 1} = 5$: 4x + 1 = 25, x = 6.
• $\sqrt[3]{x + 3} = 2$: x + 3 = 8, x = 5.
• $\sqrt{x + 12} = x$: $x^2 - x - 12 = 0$, x = 4 or −3; $\sqrt{9} = 3 \ne -3$, so x = 4 only.
• $\sqrt{x + 5} + \sqrt{x} = 5$: $x + 5 = 25 - 10\sqrt{x} + x$, $\sqrt{x} = 2$, x = 4 (3 + 2 = 5 ✓).
• $\dfrac{x - 4}{\sqrt{x} - 2} = 5$: the left side is $\sqrt{x} + 2$, so $\sqrt{x} = 3$, x = 9 (5/1 = 5 ✓).

A good set includes one question with a false root, so the class sees why checking matters.