Exercises: Radical equations

Solve the following radical equation and check the validity of the solution obtained:

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

Determine the value of $x$ satisfying the following irrational equation:

$\sqrt{2x + 7} - x = 2$

Solve the following equation containing a higher-order root:

$\sqrt[3]{5x - 4} = -1$

Find the solution to the equation by equating both radicals and check the result:

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

The time $t$ (in seconds) taken for an object to fall freely from a height $h$ (in metres) is given by the physics formula $t = \sqrt{\frac{2h}{9.8}}$. Calculate the height from which a stone was dropped if it took exactly $3$ seconds to reach the ground.

Find the values of $x$ satisfying the following equation with fractions inside the radicand:

$\sqrt{\frac{4x - 1}{3}} = 5$

The base radius $r$ of a cylinder of volume $V$ and height $h$ is calculated using the expression $r = \sqrt{\frac{V}{\pi h}}$. If a cylindrical container of height $15\text{ cm}$ has a volume of $471\text{ cm}^3$, set up the corresponding equation and find its base radius (use the approximation $\pi \approx 3.14$).

Solve the following radical equation and explain why one of the potential algebraic solutions obtained after squaring is invalid:

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

Aoife is solving the radical equation $\sqrt{3x + 1} = x - 1$ and presents the following steps on the board:

1) Squares both sides: $(\sqrt{3x + 1})^2 = (x - 1)^2 \implies 3x + 1 = x^2 - 2x + 1$.

2) Groups terms: $x^2 - 5x = 0 \implies x(x - 5) = 0$.

3) Concludes: "The solutions to the equation are $x = 0$ and $x = 5$".

1. a) Point out the conceptual error Aoife made in her conclusion and explain why the operation of squaring whole sides can introduce values that do not solve the original equation.
2. b) Write out the complete and correct resolution, carrying out a rigorous check of each candidate in the original equation to determine the final set of valid solutions.

Consider the equation with a radical $\sqrt{5x - 4} = x - 2$:

1. a) Without squaring, deduce a lower bound for the unknown taking into account that the principal square root takes non-negative values and that, therefore, $x - 2 \ge 0$ must hold. From that bound, estimate which of the integers $x = 1$, $x = 4$, or $x = 8$ can be the real solution by mentally evaluating both sides.
2. b) Solve the equation algebraically by isolating the radical and squaring both sides.
3. c) Check the obtained values in the original equation and explain how your previous estimate allowed you to discard beforehand the extraneous solution generated by the algebraic method.

Solve the following irrational equation containing two radicals:

$\sqrt{x + 4} - \sqrt{x - 1} = 1$

Calculate the $x$-coordinate where a point $P(x, 4)$ must be placed relative to point $A(-2, 1)$ so that the Euclidean distance between them is exactly $5$ units. Set up the corresponding radical equation using the distance formula and solve it.

The production cost $C$ (in thousands of euros) of manufacturing $x$ units of a high-precision electronic component follows the mathematical model $C(x) = 12 + \sqrt{5x + 4}$. If the exact budget allocated for the production of this batch is $25$ thousand euros, how many units can be manufactured?

Find the real solutions to the following irrational equation by properly isolating one of the radicals before squaring:

$\sqrt{2x - 1} + \sqrt{x + 4} = 6$

The period $T$ of oscillation of a simple pendulum (in seconds) is related to its length $L$ (in metres) by the physical expression $T = 2\pi \sqrt{\frac{L}{g}}$. Taking the acceleration due to gravity as $g \approx \pi^2\text{ m/s}^2$, set up the equation and find the length of a pendulum whose period of oscillation is exactly $2.5$ seconds.

Find the real solutions to the following equation containing a nested radical (a radical inside another):

$\sqrt{x + \sqrt{x + 3}} = 3$

Solve the following irrational equation. It will be easier if you use a suitable substitution, for example $t = \sqrt{x^2 - 2}$:

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

In an architectural project, an access ramp is designed resting against a vertical wall $12\text{ m}$ high. The length of the ramp is $\sqrt{x^2 + 144}$ and the horizontal distance from the start of the ramp to the base of the wall is $x$. If, due to safety and gradient regulations, the ramp must measure exactly $4\text{ m}$ more than the horizontal distance $x$, set up and solve the irrational equation to find $x$.

Design an equation with a radical of the form $\sqrt{a \times x + b} = x - 2$, where $a$ and $b$ are integers, that simultaneously satisfies these two conditions:

1) Upon squaring both sides and rearranging terms, the resulting quadratic equation has $x = 7$ and $x = -1$ as candidate solutions.

2) Upon checking these candidates in the original equation with the radical, $x = 7$ is a valid solution and $x = -1$ turns out to be an extraneous (false) solution.

Explain deductively how you determine the values of the coefficients $a$ and $b$ from the sum and product of the roots of the quadratic equation, write the definitive equation, and carry out the explicit verification of both values in the constructed equation.

During the modernization works of a pumping station of the Uisce Éireann (Irish Water) near Limerick on the Shannon, two flow control gates record level readings linked by the following equation with two radicals:

$\sqrt{2x + 15} - \sqrt{x - 1} = 3$

where $x$ represents the safety water depth of the reservoir in decimetres ($x \ge 1$).

1. a) Set up and solve the equation by conveniently isolating a radical and carrying out the two necessary squarings.
2. b) Check the validity of the obtained candidates by substituting them into the original equation.
3. c) Unlike what usually happens in radical equations with two squarings, analyse and justify why in this case both numerical values turn out to be valid real solutions, paying special attention to the sign of the sides just before the second squaring.

Solve the following irrational equation containing three radicals. Take special care when collecting terms after squaring for the first time:

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

A fibre-optic cable must connect two communications towers $A$ and $B$ of heights $15\text{ m}$ and $20\text{ m}$ respectively, passing through an anchor point $P$ located on the ground between both towers. The horizontal distance between the towers is $35\text{ m}$. If point $P$ is located $x$ metres from tower $A$, the total length of the cable is given by $L = \sqrt{x^2 + 225} + \sqrt{(35 - x)^2 + 400}$. If exactly $50\text{ m}$ of cable is available and is to be used in its entirety without cuts, determine at what distance $x$ the anchor $P$ should be placed.

Determine the numerical value of the parameter $k$ so that the irrational equation $\sqrt{x + k} = x - 1$ has a single valid real solution of multiplicity one. Justify your answer by analysing the discriminant of the resulting quadratic equation and the validity conditions of the radical.

Solve the following equation, which combines the properties of absolute values with irrational expressions. Carefully consider the cases for the absolute value:

$|\sqrt{3x + 1} - 4| = 2$

Solve the following irrational equation with roots of different indices. It will be very helpful to perform the substitution $t = \sqrt[6]{x}$:

$\sqrt{x} - 7\sqrt[3]{x} + 12\sqrt[6]{x} = 0$

Special relativity establishes that the relativistic mass $m$ of a moving object is given by the equation $m = \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}}$, where $m_0$ is the rest mass, $v$ is the velocity of the object and $c$ is the speed of light in a vacuum. Rearrange algebraically to find the velocity $v$ in terms of $m$, $m_0$ and $c$. Subsequently, calculate at what fraction of the speed of light $c$ a spacecraft must travel for its relativistic mass to be exactly double its rest mass.

Analyse and solve the following non-linear system of equations involving radicals. You can exploit the algebraic structure of the second equation (difference of two squares) to simplify the process:

$\begin{cases} \sqrt{x} + \sqrt{y} = 5 \\ x - y = 5 \end{cases}$

Prove algebraically and with reasoning why the irrational equation $\sqrt{x - 4} + \sqrt{2 - x} = 5$ has no solutions in the set of real numbers. Base your proof on the analysis of the domain of definition of the radicals involved, without needing to square both sides.

To solve the equation with radicals of different indices $\sqrt{2x - 1} + \sqrt[4]{2x - 1} = 6$, two classmates propose different strategies:

• Cian's method: Isolate the square root, $\sqrt{2x - 1} = 6 - \sqrt[4]{2x - 1}$, square both sides, isolate the resulting fourth root, and raise again to the fourth power to eliminate all radicals.

• Aoife's method: Define the auxiliary variable $t = \sqrt[4]{2x - 1}$, express the equation in terms of $t$, solve the resulting polynomial equation, and reverse the substitution.

1. a) Comparatively analyse both paths: state with justification the degree of the polynomial equation that Cian would obtain after his successive squarings and powers, and argue which of the two methods is more efficient and safer to avoid operational errors and high-degree extraneous roots.
2. b) Solve the equation applying Aoife's method. Justify why one of the solutions obtained for $t$ must be immediately discarded before calculating $x$, and find the unique real solution of the equation.

In the design of the braking system for a tram prototype in Dublin, the stopping distance $x$ (in metres, with $x \ge -2.5$) relative to a control beacon is regulated by the equation with an absolute value and a radical:

$|\sqrt{2x + 5} - 4| = k$

where $k$ is a physical calibration constant ($k \ge 0$).

1. a) For the particular setting $k = 1$, split the equation into two radical equations according to the definition of absolute value, solve them, and check the valid solutions for $x$.
2. b) Carry out a complete discussion of the number of real solutions that the equation has depending on the values of the parameter $k \ge 0$. Determine the intervals or exact values of $k$ for which the equation has:

• Exactly two distinct real solutions.

• Exactly one real solution.

• No real solutions.

1. c) Justify your deductions by relating the range of the principal square root $\sqrt{2x + 5} \ge 0$ to the geometric interpretation of the absolute value as distance to the value $4$ on the real number line.