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Irrational equations

This page is part of the section “Task 1 USE profile level. Elementary equations”. Here are solutions to irrational equations with neat integer roots. Each problem includes the condition, a step-by-step solution with root checking, and the answer.

Problems with solutions

Problem 1. Equation with an extraneous root

Solve the equation:

x+6=x

Square both sides. This may introduce extraneous roots, so checking is mandatory:

x+6=x2⇒x2−x−6=0

Solve the resulting quadratic equation using Vieta’s theorem:

x1=3,x2=−2

Check the roots. For x = 3: √(3 + 6) = √9 = 3 — true. For x = −2: √(−2 + 6) = √4 = 2, but the right side is −2. We get 2 = −2 — false, the root is extraneous.

Reject the extraneous root x = −2.

Answer: x=3.

Problem 2. Elementary equation

Solve the equation:

2x+3=3

Square both sides and express x:

2x+3=9⇒2x=6⇒x=3

Check: √(2·3 + 3) = √9 = 3 — true.

Answer: x=3.

Problem 3. Two roots

Solve the equation:

x2−3x+3=1

Square both sides and reduce to a quadratic equation:

x2−3x+3=1⇒x2−3x+2=0

By Vieta’s theorem the roots are:

x1=1,x2=2

Check: for x = 1 and x = 2 the radicand equals 1, √1 = 1 — both roots are valid.

Answer: x1=1, x2=2.

Problem 4. Two radicals in the equation

Solve the equation:

x+5−x−3=2

Isolate one radical on the left side:

x+5=2+x−3

Square both sides and simplify:

x+5=4+4x−3+x−3 4=4x−3⇒x−3=1⇒x−3=1⇒x=4

Check: √(4 + 5) − √(4 − 3) = √9 − √1 = 3 − 1 = 2 — true.

Answer: x=4.

Frequently asked questions

How to solve irrational equations from Task 1 of the USE?

The main method is squaring both sides. After squaring, an equation without radicals is obtained, which is solved in the standard way. A mandatory step is checking the roots: squaring may produce extraneous roots.

Why do extraneous roots appear?

Because squaring is not an equivalent transformation: if a = −b, then a² = b², but the converse is not true. Therefore, after squaring, each root must be checked by substitution into the original equation.

When can root checking be omitted?

If both sides of the equation are non-negative for all admissible x and squaring does not change the sign, checking can be replaced by the condition that the right side is non-negative. But in the USE it is safer to always check — it takes seconds.

What to do if the equation contains two radicals?

First isolate one radical (move the other to the other side), then square. After simplification, one more radical may remain — square again. Checking is mandatory at each step.

See also