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Chapter 8: Problem 44

Use the square root property to solve each equation. \((1-4 p)^{2}=24\)

Short Answer

Expert verified
The solutions are \( p = \frac{1 - 2\sqrt{6}}{4} \) and \( p = \frac{1 + 2\sqrt{6}}{4} \).

Step by step solution

01

Identify the Equation

The given equation is \( (1-4p)^2 = 24 \). Our goal is to solve for \(p\).
02

Apply the Square Root Property

To apply the square root property, take the square root of both sides of the equation. This gives us: \( \sqrt{(1-4p)^2} = \sqrt{24} \). Simplifying, we get: \( |1-4p| = \sqrt{24} \).
03

Remove the Absolute Value

Since the absolute value of \(1-4p\) equals \( \sqrt{24}\), we split it into two equations: \(1 - 4p = \sqrt{24} \) and \(1 - 4p = -\sqrt{24} \).
04

Solve for p (Case 1)

Solve the first equation \(1 - 4p = \sqrt{24} \). First, simplify \(\sqrt{24}\): \( \sqrt{24} = 2\sqrt{6} \). Then we have: \( 1 - 4p = 2\sqrt{6} \). Subtract 1 from both sides: \( -4p = 2\sqrt{6} - 1 \). Divide both sides by -4: \( p = \frac{1 - 2\sqrt{6}}{4} \).
05

Solve for p (Case 2)

Solve the second equation \(1 - 4p = -\sqrt{24} \). Using the simplified form of \(\sqrt{24}\), we have: \( 1 - 4p = -2\sqrt{6} \). Subtract 1 from both sides: \( -4p = -2\sqrt{6} - 1 \). Divide both sides by -4: \( p = \frac{1 + 2\sqrt{6}}{4} \).

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Key Concepts

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

Solving Equations
When solving equations, the main goal is to find all values of the variable that make the equation true. For most algebraic equations, this involves performing operations that simplify the equation.
These operations can include:
  • Adding or subtracting the same number from both sides
  • Multiplying or dividing both sides by the same number
  • Using properties of equality to maintain balance
For instance, to solve \( (1-4p)^2 = 24 \), we should systematically isolate the variable, in this case, 'p'. Always start by simplifying the equation as much as possible before solving for the variable.
Absolute Value Equations
Absolute value equations involve expressions within absolute value bars, meaning the distance from zero on a number line. The expression inside the absolute value brackets can be either positive or negative, but the absolute value itself is always non-negative.
For example, if we have \( |1-4p| = \sqrt{24} \), this means:
  • \(1 - 4p = \sqrt{24} \)
  • \(1 - 4p = -\sqrt{24} \)
Solving each case separately will yield all possible solutions for 'p'. It鈥檚 essential to understand that absolute value equations result in two different scenarios due to the nature of absolute values.
Radical Expressions
Radical expressions consist of roots, most often square roots, but can include other types of roots like cube roots. Simplifying these expressions often involves factoring the number under the root.
For instance, \( \sqrt{24} \) can be simplified:
  • Factor 24 into prime factors: \(24 = 2 * 12 = 2 * 2 * 6 = 2 * 2 * 2 * 3 = 2^3 * 3 \)
  • Apply the square root to the factored form: \( \sqrt{24} = \sqrt{2^3 * 3} = 2 \sqrt{6} \)
Recognizing and simplifying radical expressions helps solve equations involving roots more easily. Always simplify the radical expression before solving it with the rest of the equation.

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Most popular questions from this chapter

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A model rocket is projected vertically upward from the ground. Its distance s in feet above the ground after t seconds is given by the quadratic function $$ s(t)=-16 t^{2}+256 t $$ to see how quadratic equations and inequalities are related. At what times will the rocket be \(624 \mathrm{ft}\) above the ground? (Hint: Let \(s(t)=624\) and solve the quadratic equation.)

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