Exercises: Irrational equations

Solve the following irrational equation and check the validity of its solution:

$\sqrt{4x + 9} = 5$

Find the value of $x$ that satisfies the following equation containing a cube root:

$\sqrt[3]{3x - 5} = 2$

Determine the possible solutions to the quadratic equation under the radical:

$\sqrt{x^2 - 3x + 6} = 2$

Solve the irrational equation by isolating the radical and squaring. Pay special attention to possible false solutions:

$\sqrt{2x + 15} = x$

Find the solution by equating the radicands of the following equation:

$\sqrt{3x - 2} = \sqrt{x + 8}$

Solve for the unknown $x$ in the irrational equation containing a fraction in its radicand:

$\sqrt{\frac{5x - 1}{3}} = 3$

In kinematics, the time $t$ (in seconds) taken for an object to fall freely from a height $h$ (in metres) is given by the formula $t = \sqrt{\frac{2h}{9.8}}$. Calculate from what height a stone was dropped if it took exactly $4$ seconds to hit the ground.

The diagonal $d$ of a rectangle can be calculated from its sides $b$ and $h$ using Pythagoras' theorem: $d = \sqrt{b^2 + h^2}$. If we know that the diagonal measures $13$ cm and one of its sides measures $5$ cm, set up the corresponding irrational equation and calculate the length of the other side.

Lucas has solved the irrational equation $\sqrt{2x - 3} = x - 3$ on the board following these steps:

1) Square both sides: $2x - 3 = (x - 3)^2 \implies 2x - 3 = x^2 - 6x + 9$.

2) Collect terms on one side: $x^2 - 8x + 12 = 0$.

3) Apply the quadratic formula and obtain two solutions: $x_1 = 6$ and $x_2 = 2$.

4) Conclude: "The final solution set of the equation is $\{2, 6\}$".

1. a) Without solving the equation from scratch, identify the mistake made by Lucas and explain with reasons why the operation of squaring can introduce values that do not satisfy the original equality.
2. b) Check both values in the starting equation and determine which is the only valid real solution.

Given the irrational equation $\sqrt{2x + 1} = x - 1$:

1. a) Before operating algebraically, analyse the sign restrictions: deduce for which values of $x$ the square root exists in $\mathbb{R}$ and for which the right-hand side is non-negative. With this, delimit the interval of the real line where it is viable to find solutions.
2. b) Mentally evaluate the values $x = 0$, $x = 2$, and $x = 4$. From that evaluation, estimate between which integers the real solution must lie and argue whether there can be a solution less than $1$.
3. c) Solve the equation by squaring both sides, check whether the solution found matches your initial estimate, and formally discard any extraneous solution, justifying the reason.

Solve the equation with two radicals by appropriately isolating them on different sides before squaring:

$\sqrt{x + 5} + \sqrt{x} = 5$

Find the solutions to the following equation, making sure to check the results obtained to rule out extraneous solutions:

$x - \sqrt{2x - 1} = 2$

The period $T$ of a simple pendulum (the time it takes to complete one full oscillation) is calculated using the formula $T = 2\pi \sqrt{\frac{L}{g}}$, where $L$ is the length of the string in metres and $g = 9.8 \text{ m/s}^2$ is the acceleration due to gravity. If a pendulum clock at a science exhibition has an exact period of $2.5$ seconds, find the length of its pendulum. Use the approximation $\pi \approx 3.14$.

The distance between the points $A(2, -1)$ and $B(x, 3)$ in the Cartesian plane is exactly $5$ units. Set up the irrational equation corresponding to the distance between two points and calculate the possible coordinates $x$ of point $B$.

Solve the following equation involving the difference of two square roots. Isolate one of the roots before squaring:

$\sqrt{3x + 1} - \sqrt{2x - 1} = 1$

In an electronic components factory, the daily production cost $C(x)$ in euros as a function of $x$ units manufactured is modelled by the function $C(x) = 250 + 15\sqrt{x - 50}$. If on a given day the production cost was exactly $850$ euros, how many units were manufactured?

Solve the following irrational equation by making a suitable substitution (for example, $t = \sqrt{x}$):

$x - 7\sqrt{x} + 10 = 0$

The distance $d$ in kilometres to the horizon for an observer situated at a height $h$ in metres above sea level is approximated in simplified form by the formula $d = 3.57\sqrt{h}$. If a lookout in the crow's nest of a pirate ship spots land at a distance of $15$ km, at approximately what height is he situated?

The technical team at Irish Rail responsible for the maintenance of the overhead line on the Dublin-Cork line is studying the mechanical tension $T$ (in kilonewtons, kN) of the conductor cable as a function of its thermal expansion $x$ (in centimetres). The structural behaviour of the section follows the empirical model $T = \sqrt{8x + 17}$. To ensure proper contact with the train's pantograph, the mechanical counterweight on the mast simultaneously imposes the linear condition $T = x + 1$.

1. a) Set up the irrational equation that allows determining the expansion $x$ corresponding to the equilibrium position of the system.
2. b) Solve the equation by isolating the unknown by squaring.
3. c) Interpret physically and mathematically why one of the two algebraic solutions obtained makes no sense and must be discarded. State the permissible value of the expansion $x$ and the tension $T$ at that point.

Design an irrational equation of the form $\sqrt{ax + b} = x + c$ (where $a$, $b$, and $c$ are integers) that strictly satisfies the following three conditions:

1) When squaring both sides and combining like terms, it gives exactly the quadratic equation $x^2 - 5x - 6 = 0$.

2) The value $x = 6$ is the ONLY valid real solution of the original irrational equation.

3) The value $x = -1$ is discarded for being an extraneous solution introduced during the algebraic process.

Clearly show the reasoning that allows you to deduce the values of the coefficients $a$, $b$, and $c$, write the final equation, and verify in it the validity of $x = 6$ and the rejection of $x = -1$, justifying the sign difference.

Solve the following irrational equation containing three radicals. Make sure to check the solutions obtained at the end of the process:

$\sqrt{x + 2} + \sqrt{x - 3} = \sqrt{3x + 4}$

Find the real solutions to the equation with nested radicals:

$\sqrt{x + \sqrt{x - 2}} = 2$

Analyse with reasoning for which values of the real parameter $k > 0$ the equation $\sqrt{x + k} = x$ has exactly one valid real solution.

An engineer designs a loading ramp for a logistics warehouse. The horizontal base measures $x$ metres and the height measures $2$ metres less than the base. If the length of the ramp (the hypotenuse) is given by the expression $\sqrt{2x^2 - 4x + 4}$ and is required to measure exactly $\sqrt{34}$ metres, set up the irrational equation and determine the actual dimensions of the ramp.

Solve the following irrational equation. You can rationalise the denominator first, or cross-multiply and collect terms cleverly to facilitate the calculation:

$\frac{\sqrt{x + 3} + \sqrt{x - 3}}{\sqrt{x + 3} - \sqrt{x - 3}} = 2$

According to Einstein's Theory of Special Relativity, the Lorentz factor $\gamma$ determines time dilation and is calculated as $\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}$, where $v$ is the speed of the spacecraft and $c$ is the speed of light in a vacuum. If in a space mission time is observed to pass at $50\%$ of the Earth's rate (that is, $\gamma = 2$), solve the irrational equation to find at what fraction of the speed of light the spacecraft is travelling.

Solve the following irrational equation by strategically isolating the radicals and verifying the solutions at the end:

$\sqrt{2x + 5} - \sqrt{x - 1} = \sqrt{x + 2}$

A power line must connect a power station $A$ located on one side of a straight river $4$ km wide to a town $B$ located on the opposite side, $10$ km downstream. The cost of laying the cable underwater is $5$ million euros per km, and on land it is $3$ million euros per km. If the cable crosses the river in a straight line to a point $P$ located $x$ km downstream and then goes along land to $B$, the total cost in millions of euros is $C(x) = 5\sqrt{x^2 + 16} + 3(10 - x)$. Find the value of $x$ that makes the total cost exactly $46$ million euros by solving the corresponding irrational equation.

Consider the equation: $x^2 - 3x - \sqrt{x^2 - 3x + 5} = 1$.

Two 5th Year students propose two different solving methods:

• Strategy A (Direct squaring): Isolate the radical, $\sqrt{x^2 - 3x + 5} = x^2 - 3x - 1$, and square both sides, obtaining a 4th-degree equation in $x$ to solve it by applying Ruffini's rule.

• Strategy B (Change of variable): Take advantage of the repeated structure of the expression and define $t = \sqrt{x^2 - 3x + 5}$ (with the condition $t \ge 0$), transforming the equality into a 2nd-degree equation in $t$.

1. a) Compare both strategies with reasons, analysing the operational difficulty, the number of candidate solutions generated by each, and the risk of carrying extraneous roots. Justify which of the two paths is objectively cleaner and more efficient.
2. b) Solve the equation using Strategy B, justifying the immediate discarding of any inadmissible value of $t$ before undoing the substitution, and find all the real solutions of $x$.

In an assessment test, students are asked to discuss the number of real solutions of the equation $\sqrt{x + 5} = x + m$ as a function of the real parameter $m$.

Aoife submits the following resolution:

"I square both sides: $x + 5 = x^2 + 2mx + m^2 \implies x^2 + (2m - 1)x + (m^2 - 5) = 0$.

The discriminant is $\Delta = (2m - 1)^2 - 4(m^2 - 5) = 21 - 4m$.

Conclusion: if $m < 21/4$, since $\Delta > 0$, the irrational equation has two real solutions; if $m = 21/4$, it has one solution; and if $m > 21/4$, it has no real solutions".

1. a) Explain rigorously why Aoife's conclusion is incomplete and false, indicating which mandatory condition of irrational equations she has not taken into account when transferring the results from the quadratic to the original equation.
2. b) Provide a concrete counterexample by choosing a numerical value of $m < 21/4$ for which the irrational equation has only ONE real solution instead of two. Solve the equation with that value of $m$ and show which candidate is discarded and why.
3. c) Perform the complete and correct discussion of the number of real solutions of the equation for all possible values of $m \in \mathbb{R}$.