Problem 28 - Precalculus Student Solutions Manual 5th

Question

Solve each equation by the square root property. $$48-x^{2}=0$$

Step-by-Step Solution

Answer

x = \pm 4\sqrt{3}

Step 1 - Isolate the quadratic term

Move the quadratic term to one side of the equation: \[48 - x^2 = 0\]Add \(x^2\) to both sides: \[48 = x^2\]

Step 2 - Apply the square root property

Take the square root of both sides of the equation:\[x = \pm \sqrt{48}\]

Step 3 - Simplify the square root

Simplify the square root of 48. The square root of 48 can be written as: \[x = \pm \sqrt{16 \times 3} = \pm 4\sqrt{3}\]

Key Concepts

square root property

The square root property is a useful tool in solving quadratic equations. When you have an equation of the form Diagram 1 - Form of the equation ` x^{2}= k ` you can solve it by taking the square root of both sides. This means that : Diagram 2 - Formula to calculate Equation 1 ` x= ± k ` In simple terms, you will have two solutions: the positive square root and the negative square root. This property helps to simplify the process of solving quadratic equations by reducing them to simpler arithmetic operations.

isolating quadratic terms

Before applying the square root property, it is essential to isolate the quadratic term. This means writing the equation so that the term involving the square (i.e., Diagram 3 - Equation shows the structure ` x^{2} `) is alone on one side of the equation. In the exercise provided, we started with: `48 - x^{2}=0`. To isolate ` x^{2} `, we added ` x^{2}` to both sides, which gave us: : Diagram 4 - Form of the equation ` 48 = x^{2} ` This is crucial because it transforms the equation into a form where you can directly apply the square root property. Isolating the quadratic term is a straightforward step that paves the way for solving the equation efficiently.

simplifying square roots

Simplifying square roots involves breaking down the radicand (the number under the square root sign) into its prime factors or into the product of a perfect square and another number. This makes the square root much easier to handle. In our exercise, we ended up with `x = ± Diagram shows the way `√48 To simplify √48, we looked for factors of 48 that are perfect squares. We know that `48 = 16 * 3`, and `16` is a perfect square (since `4^2 = 16`). Therefore, ` : : ` :` ` = Diagram shows the steps `±4√3 ` This simplifies our solution giving us `x = ±4√3 `. By simplifying square roots, we can often find more manageable and meaningful expressions which help in understanding the problem and the solution better.