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Definition of radical equations with examples

radical equations

Radical equations (also known as irrational) are equations in which the unknown value appears under a radical sign.

The method for solving radical equation is raising both sides of the equation to the same power.

If we have the equation \sqrt{f(x)} = g(x), then the condition of that equation is always f(x) \geq 0, however, this is not a sufficient condition. Respecting the properties of the square root function (the domain of square root function is \mathbb{R} ^+ \cup \{0\}), the second condition is g(x) \geq 0.  Therefore, we need to ensure that both sides of equation are non-negative.

After squaring we have an equivalent equation:

    \[f(x) = [g(x)]^2.\]

Condition f(x) \geq 0 is now unnecessary (it is automatically satisfied after squaring); the solutions of the equation will thus satisfy condition g(x)  \geq 0, so that for these solutions it will be f(x) = [g(x)]^2.

Now we have:

\sqrt{f(x)} = g(x) \Leftrightarrow  g(x) \geq 0 and f(x) = [g(x)]^2.

 

In general, this is valid for the square root of every even number n:

 

 \sqrt[n]{f(x)} = g(x) \Leftrightarrow  g(x) \geq 0 and f(x) = [g(x)]^{n}.

 

For the square root of every odd number n it will be

\sqrt[n]{f(x)} = g(x) \Leftrightarrow f(x) =[g(x)]^{n},

because their domain is a whole set of real numbers.

Example 1. Solve the equation:

\sqrt{2x + 1} = 1.

Solution:

Both sides of the equation are non-negative; we can square the equation:

 

\sqrt{2x + 1} = 1 / ^2

2x + 1 = 1

2x = 0

x = 0.

We must now confirm if x = 0 it is the correct solution:

\sqrt{2\cdot 0 +1} = 1

\sqrt{1}=1

1=1.

It follows that x=0 is the solution of the given equation.

Example 2.  Solve the equation:

\sqrt{2x + 1} = \sqrt{x + 2}.

Solution:

Conditions for this equation are 2x+1 \geq 0 and x+2 \geq 0 \Rightarrow  x\geq  -\frac{1}{2} and x\geq -2.

It follows that x must be in interval [- \frac{1}{2}, + \infty  \rangle.

Both sides of the equation are always non-negative, therefore we can square the given equation.

\sqrt{2x + 1} = \sqrt{x + 2} / ^2

2x + 1 = x + 2

x = 1.

We need check that x=1 is the solution of the initial equation:

\sqrt{2\cdot 1 + 1} = \sqrt{1 + 2}

\sqrt{3}=\sqrt{3}.

 

It follows that x=1 is the solution of the initial equation. We can conclude that directly from the condition of the equation, without any further requirement to checking.

Example 3. Solve the equation:

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

Solution:

Now we must be sure that the right side of  the equation is non-negative. Therefore 2x-3 \geq 0 \Rightarrow x \geq  \frac{3}{2} is the condition of this equation.

After squaring the equation, we have:

\sqrt{x + 1} = 2x - 3  \Leftrightarrow x + 1 = 4x^2 - 12x + 9 \Leftrightarrow 4x^2 - 13x + 8 = 0.

The solutions for quadratic equation 4x^2 - 13x + 8 = 0 are:

x_1 = \frac{13 + \sqrt{41}}{8} and x_2 = \frac{13 - \sqrt{41}}{8}.

The only solution is x_1 due to satisfied condition x \geq  \frac{3}{2}.

 

Example 4. Solve the equation:

\sqrt{x + 1} = 2 + \sqrt{x + 2}

Solution:

Both sides of the equation are always non-negative, therefore we can square the equation. In this example we need to square the equation twice, as displayed below:

\sqrt{x + 1} = 2 + \sqrt{x + 2} /^2

x + 1 = 4 + 4 \sqrt{x + 2} + x + 2

4 \sqrt{x + 2} = - 5

\sqrt{x + 2} = - \frac{5}{4} /^2

x + 2 = \frac{25}{16}

x = - \frac{7}{16}.

 

 

x = - \frac{7}{16} is not the solution of the initial equation, because x \notin [-1, + \infty \rangle, which is the condition of the equation (check it!).

Note: as we observed through the steps of solving of the equation, that this equation does not have solutions before the second squaring, because the square root cannot be negative.

Example 5. Solve the equation.

\sqrt{2 \sqrt{x + 1}} = 2.

Solution:

Both sides of the equation are non-negative, therefore we can square the equation:

\sqrt{2 \sqrt{x + 1}} = 2 /^2

2 \sqrt{x + 1} = 4 /: 2

\sqrt{x + 1} = 2 /^2

x + 1 = 4

x = 3.

Let’s check that x = 3 satisfies the initial equation:

\sqrt{2 \sqrt{3+1}}=2

\sqrt{2 \sqrt{4}}=2

\sqrt{2\cdot 2}=2

\sqrt{4}=2

2=2.

It follows that x = 3 is the solution of the given equation.

Radical equations worksheets

  Solve radical equations (370.6 KiB, 144 hits)

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