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Chapter 11: Problem 42

Solve each equation. Check your solutions. $$ 4 t=\sqrt{8 t+3} $$

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01

Isolate the square root

Consider the given equation: y=4tz=\frac{\beta}{t}mult= Coordinate Because the equation contains a square root, isolate the square root term, in this case, \( \sqrt{8t + 3}\ \). Already done! The equation is: y=4t\sqrt{8t+{8t}}+{34}os=

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Key Concepts

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

Isolating Square Roots
When solving equations involving square roots, a primary goal is to isolate the square root term. This means you want the expression containing the square root to be alone on one side of the equation. Let's take a closer look at how this works with an example equation: \[ 4t = \sqrt{8t + 3} \]
To isolate the square root, follow these steps:
  • First, ensure that the equation is arranged so that all non-square root terms are on one side, and the square root term is on the opposite side.
  • In our case, this is already achieved since \( 4t \) is on the left and \( \sqrt{8t + 3} \) is on the right.
By isolating the square root, we prepare the equation for further steps like squaring both sides, which helps to eliminate the square root itself and makes solving the equation much simpler. Keeping the square root isolated is an essential preparatory step in ensuring the accuracy and simplicity of solving complex equations.
Algebraic Solutions
Algebraic manipulation is key in solving various types of equations. Once the square root is isolated, we can proceed by eliminating it through squaring both sides of the equation. For our example: \[ 4t = \sqrt{8t + 3} \]
Here's what we do next:
  • Square both sides of the equation to eliminate the square root: \[ (4t)^2 = (\sqrt{8t + 3})^2 \]
  • This gives us: \[ 16t^2 = 8t + 3 \]
  • Next, bring all terms to one side of the equation to form a standard quadratic equation: \[ 16t^2 - 8t - 3 = 0 \]
Now, we have a quadratic equation in the form \( at^2 + bt + c = 0 \). Use the quadratic formula \( t = \frac{{-b \pm \sqrt{b^2 - 4ac}}}{2a} \) for finding the roots, where:
  • a = 16
  • b = -8
  • c = -3
This step transforms any problem involving square roots into a more familiar quadratic equation, ensuring that it can be solved systematically using algebraic approaches.
Equation Checking
Verifying the solutions of an equation is crucial. It ensures that the values you obtain are actual solutions to the original problem and not extraneous solutions introduced during the solving process. For our example: \[ 4t = \sqrt{8t + 3} \]
We solve the quadratic equation \( 16t^2 - 8t - 3 = 0 \) to get potential solutions for \( t \). Suppose we got two roots, \( t_1 \) and \( t_2 \). To check these:
  • Substitute \( t_1 \) back into the original equation to see if both sides are equal:
    • \[ 4t_1 = \sqrt{8t_1 + 3} \]
  • If both sides are equal, \( t_1 \) is a valid solution. Repeat the same for \( t_2 \).

    • This process is critical because when equations involve squaring both sides, extraneous solutions can be introduced. Checking ensures that our solutions satisfy the original equation.
Thus, equation checking guarantees the accuracy of the derived solutions and is an indispensable step in the problem-solving process.

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Most popular questions from this chapter

Solve each problem. When appropriate, round answers to the nearest tenth. An object is projected directly upward from the ground. After \(t\) seconds its distance in feet above the ground is $$ s(t)=144 t-16 t^{2} $$ After how many seconds will the object be 128 ft above the ground? (Hint: Look for a common factor before solving the equation.

Solve each equation for the indicated variable. (Leave \(\pm\) in your answers.) $$ I=\frac{k s}{d^{2}} \text { for } d $$

Solve each equation for the indicated variable. (Leave \(\pm\) in your answers.) $$ R=\frac{k}{d^{2}} \text { for } d $$

Solve each equation. Check your solutions. $$ \frac{1}{x}+\frac{2}{x+2}=\frac{17}{35} $$

Solve each equation. Check your solutions. $$ \frac{6}{p}=2+\frac{p}{p+1} $$
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