Now I’m going to explain how to solve the equations with radicals, with exercises solved step by step.
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What are the equations with radicals?
The equations with radicals are those where x is within a square root.
A priori, these equations are neither first nor second degree, depending on the rest of the terms of the equation.
Let’s see what is the procedure to solve them and a few examples of equations with radicals.
How to solve an equation with radicals
The procedure for solving equations with radicals is as follows:
- Leave the term with the radical only in one member of the equation
- Square both members (this way we eliminate the root)
- Operate and generally, we will obtain a second degree equation that we will have to solve
- Determine which solution is the right one and discard the other, since it is a virtual solution (we have forced it by squaring the members, but it does not exist in reality)
Solved exercise of equations with radicals step by step
I am going to explain to you now more slowly how to solve the equations with radicals, following the previous procedure while we solve some exercises.
We start by leaving the root alone in one of the limbs:
We squared both limbs:
In the first member, the root is annulled with the square and in the second member there remains a remarkable product to be developed:
We are left with a second-degree equation. To solve it, we pass all the terms to the first member, leaving zero in the second member:
We solve the equation with the general formula to solve second degree equations:
We obtain two solutions, which are:
The two solutions are not valid, since one of them was created by us squaring both members, a square that did not exist in the original equation.
To find which solution is valid, we have to substitute the value of both solutions in the original equation and check which of them meets the equality.
The iguladadad is fulfilled
Equality is not fulfilled.
Therefore, the solution of the equation is x=4.
Let’s see another example of an equation with radicals with decimal numbers. We are used to the fact that both the numbers and the solutions of the equations are integers, but it doesn’t have to be that way.
Let’s see it. For example:
We leave the root term only on one of the members:
I remind you that we have added this step, that is, we are adding a solution that does not really exist.
We develop the squares in both the first and the second member:
To arrive at this result we must have clear properties of the powers, as well as notable products.
We have a complete second degree equation. To solve it, we reorder terms and pass them all to a member of the equation to equal zero:
Both members have the same result. Then this solution is valid.
With x= 2948,73:
Here, every member has a result, therefore, this solution is not valid.
The correct result is 10307,8.
How to solve equations with two radicals
So far we have seen how to solve equations with radicals, in which we had only one term with radicals.
But how do you solve equations with two radicals?
To solve the equations with two radicals, we must repeat the process 2 times, in order to eliminate the two radicals.
At the end, we must check which solution is valid, just like for equations with a radical.
Let’s see an example:
When we have two radicals in an equation, we must begin by leaving one of the races in only one of the members.
In this equation we already have one radical only, so we squared both members:
In the first member the root is annulled with the square and in the second, we develop the remarkable product, where one of the terms is the other root:
We operate on every term. When the root is squared in the remarkable product disappears:
We are left with an equation where we only have one radical, so we have to repeat the process once again.
We leave the root alone in the second limb:
We simplify terms:
We squared both terms:
In the first term we develop the remarkable product. In the second term we squared both factors of the parenthesis:
We eliminate the parenthesis of the second member by multiplying both terms by 16:
We have a second-degree equation. To solve it, we pass all terms to one member and equal zero:
We apply the general formula for solving second-degree equations:
And we get two solutions:
Let’s check which of the two is valid.
For x=61, we substitute in the original equation and operate:
Equality is not met, so x=61 is not a valid solution.
Equality is met, so x=5 is the solution to the equation.