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1st BTL CCSS · Equations and systems Notes 7 Nov 2025

6. Irrational equations

The strategy for solving equations with square roots (isolate + square + check), a worked example with three roots (which ends with no solution) and 10 exercises (36a–f and 37a–d) with the solution hidden.

Idea: isolate the root and square both sides

Irrational equation

We call irrational equations those that have the unknown inside a square root (or, more generally, inside a radical).

For example: $\;\sqrt{2x-1} + \sqrt{x-3} = \sqrt{5-x}\;$ or $\;3 + \sqrt{x^{2}-3} = x^{2}$.

General strategy (3 steps)

1) Isolate one root on one of the two sides of the equation.

2) Square both sides to make it disappear. If roots still remain (because we had two or three), isolate and square again.

3) Solve the resulting polynomial equation and check every solution in the original equation.

Notable products we'll use often

$$(a+b)^{2} = a^{2} + 2ab + b^{2}, \qquad (a-b)^{2} = a^{2} - 2ab + b^{2}.$$

When we square an expression with a root added to (or subtracted from) another term, the cross term $\;2ab$ — with a root inside — always appears.

Mandatory check — extraneous solutions

Squaring is not a reversible operation: it can create solutions that do not satisfy the original equation (we call them extraneous, just as with rational equations). In particular, if isolating a root leaves $\;\sqrt{\,\cdot\,} = E\;$ with $\,E < 0$, that branch yields no valid solution.

You must always substitute the candidate values into the original equation. If one does not fit, it is discarded.

Worked example (three roots, no solution)

$\sqrt{2x-1} + \sqrt{x-3} = \sqrt{5-x}$

We have three roots; we'll need to square twice. We start by squaring directly:

$$\bigl(\sqrt{2x-1} + \sqrt{x-3}\bigr)^{2} = \bigl(\sqrt{5-x}\bigr)^{2}.$$
$$(2x-1) + 2\sqrt{(2x-1)(x-3)} + (x-3) = 5 - x.$$

We isolate the remaining root and group terms:

$$2\sqrt{(2x-1)(x-3)} = 5 - x - 2x + 1 - x + 3 = 9 - 4x.$$

We square again:

$$\bigl(2\sqrt{(2x-1)(x-3)}\bigr)^{2} = (9-4x)^{2}.$$
$$4(2x-1)(x-3) = 81 - 72x + 16x^{2}.$$

We expand the left-hand side: $\;(2x-1)(x-3) = 2x^{2} - 7x + 3$, so $\;4(2x^{2}-7x+3) = 8x^{2} - 28x + 12$. Equating:

$$8x^{2} - 28x + 12 = 16x^{2} - 72x + 81 \;\Longrightarrow\; 0 = 8x^{2} - 44x + 69.$$

Discriminant: $\;\Delta = 44^{2} - 4 \cdot 8 \cdot 69 = 1936 - 2208 = -272 < 0$. The quadratic equation has no real root, and therefore neither does the original equation.

$$\boxed{\;\text{No solution.}\;}$$

Here it is the final quadratic itself that tells us there is no solution — we don't even need to check.

Exercises

36 Solve the following irrational equations

Afterwards, check your results with GeoGebra.

a) $x + \sqrt{x+1} = 2x - 1$

Isolate the root: $\;\sqrt{x+1} = x - 1$. Square both sides:

$$x + 1 = (x-1)^{2} = x^{2} - 2x + 1 \;\Longrightarrow\; 0 = x^{2} - 3x = x(x-3).$$

Candidates: $x = 0$ or $x = 3$.

Check: $x = 0$: $\;0 + \sqrt{1} = 1 \neq -1$ → discard.  $x = 3$: $\;3 + \sqrt{4} = 5 = 2\cdot 3 - 1$ ✓.

$$\boxed{\;x = 3.\;}$$
b) $3 + \sqrt{x^{2}-3} = x^{2}$

Isolate the root: $\;\sqrt{x^{2}-3} = x^{2} - 3$. Square both sides:

$$x^{2} - 3 = (x^{2})^{2} - 2 \cdot x^{2} \cdot 3 + 9 = x^{4} - 6x^{2} + 9.$$
$$0 = x^{4} - 7x^{2} + 12.$$

It's a biquadratic. Substitute $\,x^{2} = t$:

$$0 = t^{2} - 7t + 12 = (t-3)(t-4) \;\Longrightarrow\; t = 3 \text{ or } t = 4.$$

Undoing the substitution: $\;x = \pm\sqrt{3}\,$ or $\,x = \pm 2$.

Check: all four values satisfy $x^{2}-3 \geq 0$ and $x^{2}-3 \geq 0$ (the right-hand side after isolating). We verify: $\,3 + \sqrt{0} = 3 = (\pm\sqrt{3})^{2}\,$ ✓  and  $\,3 + \sqrt{1} = 4 = (\pm 2)^{2}\,$ ✓.

$$\boxed{\;x = \pm\sqrt{3}\ \text{or}\ x = \pm 2.\;}$$
c) $\sqrt{x} + \sqrt{x+1} = 1$

Isolate one root: $\;\sqrt{x+1} = 1 - \sqrt{x}$. Square both sides:

$$x + 1 = 1 - 2\sqrt{x} + x \;\Longrightarrow\; 0 = -2\sqrt{x} \;\Longrightarrow\; \sqrt{x} = 0 \;\Longrightarrow\; x = 0.$$

Check: $\sqrt{0} + \sqrt{1} = 0 + 1 = 1$ ✓.

$$\boxed{\;x = 0.\;}$$
d) $\sqrt{7-3x} - x = 7$

Isolate the root: $\;\sqrt{7-3x} = x + 7$. Square both sides:

$$7 - 3x = (x+7)^{2} = x^{2} + 14x + 49 \;\Longrightarrow\; 0 = x^{2} + 17x + 42.$$
$$x = \dfrac{-17 \pm \sqrt{289 - 168}}{2} = \dfrac{-17 \pm 11}{2} \;\Rightarrow\; x = -3\ \text{or}\ x = -14.$$

Check: $x = -3$: $\;\sqrt{16} - (-3) = 4 + 3 = 7$ ✓.  $x = -14$: $\;\sqrt{49} - (-14) = 7 + 14 = 21 \neq 7$ → discard.

$$\boxed{\;x = -3.\;}$$
e) $3\sqrt{6x+1} - 5 = 2x$

Isolate the root: $\;3\sqrt{6x+1} = 2x + 5$. Square both sides:

$$9(6x+1) = (2x+5)^{2} \;\Longrightarrow\; 54x + 9 = 4x^{2} + 20x + 25.$$
$$0 = 4x^{2} - 34x + 16 \;\Longrightarrow\; 0 = 2x^{2} - 17x + 8.$$
$$x = \dfrac{17 \pm \sqrt{289 - 64}}{4} = \dfrac{17 \pm 15}{4} \;\Rightarrow\; x = 8\ \text{or}\ x = \tfrac{1}{2}.$$

Check: $x = 8$: $\;3\sqrt{49} - 5 = 21 - 5 = 16 = 2\cdot 8$ ✓.  $x = \tfrac{1}{2}$: $\;3\sqrt{4} - 5 = 6 - 5 = 1 = 2\cdot\tfrac{1}{2}$ ✓.

$$\boxed{\;x = 8\ \text{or}\ x = \tfrac{1}{2}.\;}$$
f) $\sqrt{x+4} = 3 - \sqrt{x-1}$

Square both sides directly:

$$x + 4 = 9 - 6\sqrt{x-1} + (x-1) = 8 + x - 6\sqrt{x-1}.$$

Isolate the remaining root: $\;-4 = -6\sqrt{x-1} \;\Rightarrow\; \sqrt{x-1} = \dfrac{2}{3}$. Square both sides:

$$x - 1 = \dfrac{4}{9} \;\Longrightarrow\; x = \dfrac{13}{9}.$$

Check: $\sqrt{\tfrac{13}{9}+4} = \sqrt{\tfrac{49}{9}} = \tfrac{7}{3}$,  $3 - \sqrt{\tfrac{13}{9}-1} = 3 - \tfrac{2}{3} = \tfrac{7}{3}$ ✓.

$$\boxed{\;x = \dfrac{13}{9}.\;}$$
37 Solve the following equations

Check your results with GeoGebra.

a) $x + \sqrt{x+1} - 2x = 1$

Simplify: $\;-x + \sqrt{x+1} = 1 \;\Rightarrow\; \sqrt{x+1} = x + 1$. Square both sides:

$$x + 1 = (x+1)^{2}.$$

Let $u = x+1$: $\;u = u^{2} \Rightarrow u(u-1) = 0 \Rightarrow u = 0$ or $u = 1$. So $\,x = -1$ or $\,x = 0$.

Check: $x=-1$: $\;-1 + \sqrt{0} - (-2) = -1 + 0 + 2 = 1$ ✓.  $x=0$: $\;0 + \sqrt{1} - 0 = 1$ ✓.

$$\boxed{\;x = -1\ \text{or}\ x = 0.\;}$$
b) $\sqrt{x-3} + \sqrt{x+4} = \sqrt{4x+1}$

Square both sides directly:

$$(x-3) + 2\sqrt{(x-3)(x+4)} + (x+4) = 4x + 1.$$

Isolate the root: $\;2\sqrt{(x-3)(x+4)} = 4x + 1 - 2x - 1 = 2x \;\Rightarrow\; \sqrt{(x-3)(x+4)} = x$. Square both sides:

$$(x-3)(x+4) = x^{2} \;\Longrightarrow\; x^{2} + x - 12 = x^{2} \;\Longrightarrow\; x = 12.$$

Check: $\;\sqrt{9} + \sqrt{16} = 3 + 4 = 7 = \sqrt{49}$ ✓.

$$\boxed{\;x = 12.\;}$$
c) $x - \sqrt{x^{2}-3} = \sqrt{x-1}$

Square both sides:

$$\bigl(x - \sqrt{x^{2}-3}\bigr)^{2} = x - 1 \;\Longrightarrow\; x^{2} - 2x\sqrt{x^{2}-3} + (x^{2}-3) = x - 1.$$

Isolate the root: $\;2x^{2} - 3 - x + 1 = 2x\sqrt{x^{2}-3} \;\Rightarrow\; 2x^{2} - x - 2 = 2x\sqrt{x^{2}-3}$. Square both sides:

$$(2x^{2} - x - 2)^{2} = 4x^{2}(x^{2} - 3).$$
$$4x^{4} - 4x^{3} - 7x^{2} + 4x + 4 = 4x^{4} - 12x^{2}.$$
$$0 = 4x^{3} - 5x^{2} - 4x - 4.$$

By synthetic division (Ruffini) we try $x = 2$: $\,32 - 20 - 8 - 4 = 0$ ✓. Then $\,4x^{3} - 5x^{2} - 4x - 4 = (x-2)(4x^{2} + 3x + 2)$, and the quadratic has $\Delta = 9 - 32 < 0$, with no further real roots.

Check: $x = 2$: $\;2 - \sqrt{1} = 1 = \sqrt{1}$ ✓.

$$\boxed{\;x = 2.\;}$$
d) $\sqrt{x} + \dfrac{1}{\sqrt{x}} = \dfrac{x+1}{2}$

Multiply everything by $\,2\sqrt{x}\,$ (condition $x > 0$):

$$2x + 2 = \sqrt{x}\,(x+1).$$

Square both sides:

$$(2x+2)^{2} = x\,(x+1)^{2} \;\Longrightarrow\; 4x^{2} + 8x + 4 = x^{3} + 2x^{2} + x.$$
$$0 = x^{3} - 2x^{2} - 7x - 4.$$

By synthetic division (Ruffini) we try $x = 4$: $\,64 - 32 - 28 - 4 = 0$ ✓. Then $\,x^{3} - 2x^{2} - 7x - 4 = (x-4)(x^{2} + 2x + 1) = (x-4)(x+1)^{2}$. Candidates: $x = 4$ and $x = -1$ (double).

Check: $x = -1$: $\sqrt{-1}$ is undefined → discard.  $x = 4$: $\;\sqrt{4} + \tfrac{1}{\sqrt{4}} = 2 + \tfrac{1}{2} = \tfrac{5}{2} = \tfrac{4+1}{2}$ ✓.

$$\boxed{\;x = 4.\;}$$