Problem 44 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 1-50: Solve the equation. $$ 8 t-22 t^{1 / 2}-21=0 $$

Step-by-Step Solution

Answer

The solution to the equation is \( t = 12.25 \).

Step 1: Substitute to simplify

Let us substitute a new variable, \( u \), where \( u = t^{1/2} \). Therefore, \( u^2 = t \). The equation \( 8t - 22t^{1/2} - 21 = 0 \) can be rewritten in terms of \( u \) as: \[ 8u^2 - 22u - 21 = 0 \]

Step 2: Use the quadratic formula

The quadratic equation \( 8u^2 - 22u - 21 = 0 \) is of the form \( au^2 + bu + c = 0 \). Here \( a = 8 \), \( b = -22 \), and \( c = -21 \). The quadratic formula is given by \[ u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Substitute the given values into the formula.

Step 3: Calculate the discriminant

First, calculate the discriminant \( b^2 - 4ac \): \[ b^2 - 4ac = (-22)^2 - 4 imes 8 imes (-21) \] Calculate: \[ 484 + 672 = 1156 \] The discriminant is \( 1156 \).

Step 4: Solve for \( u \)

Now substitute the discriminant back into the quadratic formula: \[ u = \frac{-(-22) \pm \sqrt{1156}}{2 imes 8} \] Simplify further:\[ u = \frac{22 \pm 34}{16} \] This gives two solutions for \( u \):1. \( u = \frac{22 + 34}{16} = \frac{56}{16} = 3.5 \) 2. \( u = \frac{22 - 34}{16} = \frac{-12}{16} = -0.75 \)

Step 5: Transform back to \( t \)

Recall \( u = t^{1/2} \), so we need \( u \geq 0 \) which means discard \( u = -0.75 \). Now calculate \( t \) for \( u = 3.5 \): \[ t = u^2 = (3.5)^2 = 12.25 \]

Key Concepts

Substitution Method

The substitution method is a technique used to transform an equation into a simpler form. It's especially useful in problems that involve non-linear terms, like roots or powers that are difficult to manage directly. In this problem, we start with an equation involving a square root, specifically, \( t^{1/2} \). This can be cumbersome to deal with, so we introduce a substitution.

We set \( u = t^{1/2} \) to simplify the equation. Consequently, \( u^2 = t \). Now, the challenging root is replaced with the variable \( u \), allowing us to rewrite the equation as a quadratic: \( 8u^2 - 22u - 21 = 0 \). This substitution makes it easier to solve using standard techniques for quadratic equations.

• By substituting, we reduce complexity.
• We open the path to apply the quadratic formula.
• It's crucial to convert back to the original variable at the end.

Quadratic Formula

The quadratic formula is a universal tool for solving quadratic equations of the form \( ax^2 + bx + c = 0 \). It's applicable here because the substitution method brought us to the equation \( 8u^2 - 22u - 21 = 0 \).

The quadratic formula helps find the solutions for \( u \): \[ u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] For this equation, we substitute:

• \( a = 8 \)
• \( b = -22 \)
• \( c = -21 \)

This formula lets you calculate both potential values of \( u \), effectively solving the quadratic equation. Its reliance on the coefficients \( a \), \( b \), and \( c \) makes it a flexible and dependable solution method.

Discriminant Calculation

Calculating the discriminant is an essential part of using the quadratic formula, as it determines the nature of the roots of a quadratic equation. The discriminant \( \Delta \) is given by \( b^2 - 4ac \) and provides valuable information.

In our exercise, we calculated: \[ \Delta = (-22)^2 - 4 \times 8 \times (-21) \]Performing this calculation step-by-step, we get:

• \( 484 \) from \((-22)^2\)
• \( +672 \) since \(-4 \times 8 \times -21 = 672\)

This results in a discriminant of \( 1156 \).

• If the discriminant is positive, as it is here, it indicates two real and distinct solutions.
• If it were zero, there would be exactly one real solution.
• If it were negative, the solutions would be complex numbers.

Variable Transformation

Variable transformation is crucial in bringing the problem back to its original context, particularly after solving for the substitute variable. After finding solutions for \( u \), we need to express these results in terms of \( t \), our original variable.

From the quadratic formula, we derived two solutions for \( u \): \( 3.5 \) and \(-0.75 \). However, since \( u = t^{1/2} \), and the square root must be non-negative in real numbers, we disregard \( u = -0.75 \). To revert back to \( t \), we use \( t = u^2 \). For the valid solution \( u = 3.5 \):

• Calculate \( t = 3.5^2 \).
• Hence, \( t = 12.25 \).

The transformation is a vital step to ensure that our solution applies to the original variable's context, giving us the final answer to the problem.