Irrational equations
You already know equations with radicals from the previous section. Here the focus changes: it is not a catalogue of forms, but the discipline of checking. Every time you square both sides you may gain solutions that did not exist; in Bachillerato you will be asked to justify rigorously why a solution is accepted or rejected, and even to reason about when an irrational equation has no solution without solving it.
The method: isolate, square and always check
Solve the following irrational equation and check the validity of its solution: $\sqrt{4x + 9} = 5$
Solve the following irrational equation and check the validity of its solution:
$\sqrt{4x + 9} = 5$
Solve the irrational equation by isolating the radical and squaring. Pay special attention to possible false solutions: $\sqrt{2x + 15} = x$
Solve the irrational equation by isolating the radical and squaring. Pay special attention to possible false solutions:
$\sqrt{2x + 15} = x$
More elaborate radicands
Find the value of $x$ that satisfies the following equation containing a cube root: $\sqrt[3]{3x - 5} = 2$
Find the value of $x$ that satisfies the following equation containing a cube root:
$\sqrt[3]{3x - 5} = 2$
Solve for the unknown $x$ in the irrational equation containing a fraction in its radicand: $\sqrt{\frac{5x - 1}{3}} = 3$
Solve for the unknown $x$ in the irrational equation containing a fraction in its radicand:
$\sqrt{\frac{5x - 1}{3}} = 3$
Two radicals: squaring twice
Solve the equation with two radicals by appropriately isolating them on different sides before squaring: $\sqrt{x + 5} + \sqrt{x} = 5$
Solve the equation with two radicals by appropriately isolating them on different sides before squaring:
$\sqrt{x + 5} + \sqrt{x} = 5$
Solve the following equation involving the difference of two square roots. Isolate one of the roots before squaring: $\sqrt{3x + 1} - \sqrt{2x - 1} = 1$
Solve the following equation involving the difference of two square roots. Isolate one of the roots before squaring:
$\sqrt{3x + 1} - \sqrt{2x - 1} = 1$
Change of variable
Solve the following irrational equation by making a suitable substitution (for example, $t = \sqrt{x}$): $x - 7\sqrt{x} + 10 = 0$
Solve the following irrational equation by making a suitable substitution (for example, $t = \sqrt{x}$):
$x - 7\sqrt{x} + 10 = 0$
Parameter and domain: reasoning without solving
Analyse with reasoning for which values of the real parameter $k > 0$ the equation $\sqrt{x + k} = x$ has exactly one valid real solution.
Analyse with reasoning for which values of the real parameter $k > 0$ the equation $\sqrt{x + k} = x$ has exactly one valid real solution.
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…
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.
Applications of irrational equations
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…
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.
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…
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 $d$ 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…
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?