# Question Video: Finding the Solution Set of Root Equations Involving Absolute Value Mathematics • 9th Grade

Find the solution set of the equation √(4𝑥² − 28𝑥 + 49) = |𝑥 + 4|.

03:52

### Video Transcript

Find the solution set of the equation the square root of four 𝑥 squared minus 28𝑥 plus 49 is equal to the absolute value of 𝑥 plus four.

In this question, we’re asked to find the solution set of a given equation involving radicals and the absolute value. To do this, we first recall the solution set is the set of all solutions to this equation. So we need to start by solving this equation. That’s finding all the values of 𝑥 such that the left-hand side is equal to the right-hand side of the equation. So to solve this equation, let’s start by taking a look at the equation.

On the left-hand side, we have the square root of four 𝑥 squared minus 28𝑥 plus 49. And usually, the easiest way to solve an equation involving a square root is to square both sides of the equation. However, if we do this, we can introduce extra solutions. So we do need to be careful. This is particularly important because we’re taking the square root of a number. For example, if we take the square root of a negative number, we’ll end up with a complex number.

However, there is one interesting thing worth noting about this example. On the left-hand side of this equation, we’re taking the square root, which means we take the positive value. And on the right-hand side, we’re taking the absolute value. So this is also the positive value. So both sides of the equation are already positive. This can help justify taking the squares of both sides of the equation. This gives us that four 𝑥 squared minus 28𝑥 plus 49 is equal to the absolute value of 𝑥 plus four squared.

And we can simplify the right-hand side of this equation by remembering the absolute value of a number is its magnitude. It doesn’t matter about the sign. But if we’re squaring this value, it doesn’t matter if we take the positive or negative value. In other words, for any real number 𝑎, the magnitude of 𝑎 squared is just equal to 𝑎 squared. Therefore, we can use this to simplify the right-hand side of our equation to be 𝑥 plus four all squared.

And now, we can simplify the right-hand side of this equation further by distributing the exponent over the parentheses. We can do this by using the FOIL method or binomial expansion. In either case, we get 𝑥 squared plus eight 𝑥 plus 16. And remember, this needs to be equal to four 𝑥 squared minus 28𝑥 plus 49. And now, we’re just solving a quadratic equation. We can do this by collecting like terms. So we subtract 𝑥 squared from both sides of the equation, eight 𝑥 from both sides of the equation, and 16 from both sides of the equation and simplify. We get three 𝑥 squared minus 36𝑥 plus 33 is equal to zero.

And now, there’s many different ways of solving a quadratic equation. We’re going to start by noticing all three terms share a factor of three. So we can just divide through by three. This gives us 𝑥 squared minus 12𝑥 plus 11 is equal to zero. And we can solve this by factoring. We need two numbers which multiply to give 11 and add to give negative 12. And of course, negative one times negative 11 is equal to 11, and negative one plus negative 11 is equal to negative 12.

Therefore, we can factor this quadratic to get 𝑥 minus one multiplied by 𝑥 minus 11 is equal to zero. And finally, for the product of two numbers to be equal to zero, one of the two factors must be equal to zero. In other words, either 𝑥 is equal to one or 𝑥 is equal to 11. And in this case, it’s not actually necessary to check whether these two solutions are valid or not because, as we already said, both sides of the equation were already positive. And if we substituted these values into the left-hand side of that equation and got a negative value, there wouldn’t be a solution because the right-hand side of the equation is already positive.

So by being solutions to the equation, we guarantee that the radical is defined at these values of 𝑥. However, this won’t always be the case for all equations of this form. So it can be a good idea to check anyway. We would just substitute 𝑥 is equal to one and 𝑥 is equal to 11 into our original equation to check that they are indeed solutions. And in both cases, we would determine that they are indeed solutions.

Therefore, we were able to determine the solution set of the equation the square root of four 𝑥 squared minus 28𝑥 plus 49 is equal to the absolute value of 𝑥 plus four is the set containing one and 11.