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Solve the equations using any method you choose. $$x^{2}=24$$

Short Answer

Expert verified
The solutions are \(x = \pm 2\sqrt{6}\).

Step by step solution

01

Set Up the Equation

We are given the equation \(x^2 = 24\). Our goal is to solve for \(x\).
02

Use Square Root Property

To solve for \(x\), we can use the square root property, which states that if \(x^2 = a\), then \(x = \pm \sqrt{a}\). Therefore, apply the square root to both sides of the equation: \(x = \pm \sqrt{24}\).
03

Simplify the Square Root

Next, simplify \(\sqrt{24}\). First, factor 24 into its prime factors: \(24 = 2^3 \cdot 3\). We can simplify \(\sqrt{24}\) as follows: \(\sqrt{24} = \sqrt{2^3 \cdot 3} = 2\sqrt{6}\).
04

Write Final Solution

Using the simplified form of the square root from the previous step, the solution to the equation is \(x = \pm 2\sqrt{6}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Square Root Property
The square root property is a vital concept in algebra, especially when solving quadratic equations. It provides an efficient way to find the roots of an equation where the variable is squared. This property states that for any equation of the form \(x^2 = a\), the solutions for \(x\) are \(x = \pm \sqrt{a}\). The reasoning behind this is that squaring both \(\sqrt{a}\) and \(-\sqrt{a}\) yields \(a\), thus both are solutions. This property is useful in cases where you can easily isolate the squared term on one side of the equation.
To use this property effectively, make sure the equation is simplified such that the quadratic term is isolated on one side of the equation and equated to a constant or another expression. Then apply the property to find the possible values of the variable. Remember, the solution will typically involve both the positive and negative roots, reflecting the fact that two numbers can yield the same result when squared.
Simplifying Square Roots
Simplifying square roots is an important step in obtaining the cleanest form of a solution. Many numbers that are not perfect squares can still be broken down into simpler components. The process involves expressing the number inside the square root as a product of factors and, if possible, identifying perfect squares within these factors.
  • Start by finding the prime factorization of the number, as this can help you see which factors can be taken out of the square root.
  • Look for pairs of identical factors, as these represent perfect squares and can be simplified.
  • Move any factor that forms a perfect square out of the square root as its square root.
For example, the number 24 can be factored as \(2^3 \cdot 3\). Since \(2^2 = 4\) is a perfect square, we can extract 2 outside the square root, simplifying \(\sqrt{24}\) to \(2\sqrt{6}\). This step helps to achieve the most simplified and readable form of the solution.
Solving Equations
Solving equations is a fundamental skill in mathematics that involves finding the value of the variable that makes the equation true. For quadratic equations, you often use methods like factoring, completing the square, or using the quadratic formula. In this particular exercise, we utilized the square root property as it provides a straightforward means of solving equations of the form \(x^2 = a\).
When dealing with equations:
  • Always aim to isolate the variable you are solving for. This might involve moving terms across the equal sign and simplifying the expression.
  • Consider the nature of the equation—which solving technique best suits it? In cases where the quadratic expression has no linear term, the square root method is effective.
  • After finding the potential solutions, substitute them back into the original equation to ensure they satisfy the equation. This step helps verify the accuracy of your solutions.
Using these steps ensures that the solution is as precise as possible and confirms its validity within the context of the given equation.

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Most popular questions from this chapter

You \(\%\) need to recall the following definitions and results from elementary geometry. In a triangle, a line segment drawn from a vertex to the midpoint of the opposite side is called a median. The three medians of a triangle are concurrent; that is, they intersect in a single point. This point of intersection is called the centroid of the triangle. A line segment drawn from a vertex perpendicular to the opposite side is an altitude. The three altitudes of a triangle are concurrent; the point where the altitudes intersect is the orthocenter of the triangle. This exercise provides an example of the fact that the medians of a triangle are concurrent. (a) The vertices of \(\triangle A B C\) are as follows: $$A(-4,0) \quad B(2,0) \quad C(0,6)$$ Use a graphing utility to draw \(\triangle A B C\). (since \(\overline{A B}\) coincides with the \(x\) -axis, you won't need to draw a line segment for this side.) Note: If the graphing utility you use does not have a provision for drawing line segments, you will need to determine an equation for the line in each case and then graph the line. (b) Find the coordinates of the midpoint of each side of the triangle, then include the three medians in your picture from part (a). Note that the three medians do appear to intersect in a single point. Use the graphing utility to estimate the coordinates of the centroid. (c) Using paper and pencil, find the equation of the medians from \(A\) to \(\overline{B C}\) and from \(B\) to \(\overline{A C}\). Then (using simultaneous equations from intermediate algebra), determine the exact coordinates of the centroid. How do these numbers compare with your estimates in part (b)?

Explain why there are no real numbers that satisfy the equation \(\left|x^{2}+4 x\right|=-12\).

Find an equation for the line passing through the two given points. Write your answer in the form \(y=m x+b\). (a) (4,8) and (-3,-6) (b) (-2,0) and (3,-10) (c) (-3,-2) and (4,-1)

Show that the slope of the line passing through the two points (3,9) and \(\left(3+h,(3+h)^{2}\right)\) is \(6+h\).

Find an equation for the line having the given slope and passing through the given point. Write your answers in the form \(y=m x+b\). (a) \(m=-5 ;\) through (-2,1) (b) \(m=\frac{1}{3} ;\) through \(\left(-6,-\frac{2}{3}\right)\)

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