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Equations involving a radical in practice

Equations involving a radical in practice
10.5
Hello and welcome back to a step in practice. We are dealing here with equations involving a radical. In exercise 1, we want to solve the equation 3 plus square of x minus 1 equals x.
34.1
In every equation, we have first of all to study the domain. The domain here is the set of numbers, such that x minus 1 is greater or equal than 0. Otherwise, the square root of x minus 1 does not make sense. So it’s the interval 1 plus infinity. So we write here that we need that x be greater than 1.
64.8
We must remember this condition.
69.5
Now, we put the square on the left-hand side, and the linear terms on the right-hand side, and we write that the equation is equivalent when x is in D. So this was the first step, this is the second step. For x in D, the equation is equivalent to square root of x minus 1, equals x minus 3. And now recall what Francis told us studying this kind of equations. This is equivalent to two conditions. x minus 3 greater or equal than 0, and the squared the equation, x minus 1, which is square root of x minus 1 to the square equals x minus 3 to the square. So the first condition gives us x greater or equal than 3.
131.8
So another constraint, and we have to take care and remember it. Now we studied the squared equation. So x minus 1 equals x minus 3 to the square is equivalent to x minus 1 equals x squared minus 6x plus 9, and this turns out to be equivalent to x squared minus 7x plus 10 equal to 0. Now, the discriminant of the equation is 49 minus 40. It is 9, 3 to the square. And thus, there are two roots, 1 is equal equal– x equals 7 minus 3 over 2, or x equals 7 plus 3 over 2. This gives 2 or x equals 5. Now we look at the compatibility with the other conditions that we found.
212.2
We must have x greater than 3 and x greater than 1, that is, x greater than 3, so we exclude 2 and it remains x equals 5. So the solution is 5. Let us check that it works. Square root of 5 minus 1 is square root of 4, that is 2, and 2 plus 3 gives 5. And it is true that when x is 5, so the equality is satisfied.
246.7
In exercise 2, we are asked to solve the equation 2x plus square root of 2 minus x equals 1. Again, we study the natural domain of the equation, which is the set D equals to the x satisfying 2 minus x greater or equal than 0 because we want the square root of 2 minus x to be defined. So this means the interval minus infinity 2. So take care. And what follows, we must be sure that, at the end, x will be less or equal than 2. Let us write it here, apart.
298.4
Now, for x in D, the equation is equivalent to square root of 2 minus x equals to 1 minus 2x, and we know as Francis showed us, that this is equivalent to a couple of facts. One, 1 minus 2x is greater or equal than 0, and 2 minus x, we square the equation 2 minus x equals 1 minus 2x to the square. Now, take care. This implies that x should be less or equal than 1/2. So we write here x less or equal than 1/2.
347.6
And now we studied the squared equation.
352.2
So it is equivalent to 2 minus x equals 1 plus 4x to the square minus 4x, which is 4 x square minus 3x minus 1 equals to 0. Now, you see there is one evident root, x equal to 1, 4 minus 3 minus 1 gives 0.
382.5
And the other root, it’s minus 1 over 4.
391.6
We can reach the result also by studying the discriminant, by means of the formulas. Now, we study the compatibility of the solutions with our conditions. We know that x must be less or equal than 1/2. So of course, we must exclude x equal to 1. It remains x equal to minus 1/4, which is actually less than 2 and less than 1/2. So the solution is minus 1/4.
428.5
And this ends our step.
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See you in the next step.

The following exercises are solved in this step.

We invite you to try to solve them before watching the video.

In any case, you will find below a PDF file with the solutions.

Exercise 1.

Solve the equation (3+sqrt{x-1}=x).

Exercise 2.

Solve the equation (2x+sqrt{2-x}=1).

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Precalculus: the Mathematics of Numbers, Functions and Equations

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