Chapter 21: Chapter 21

Problem 631

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Solve the equation \(4 \mathrm{x}^{3}-24 \mathrm{x}^{2}+23 \mathrm{x}+18=0\), having given that the roots are in arithmetical progression.

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The roots of the given equation \(4x^3 - 24x^2 + 23x +18 = 0\) are -4, -2, and 0.

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TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Set up the roots

Let the roots of the given equation be r-d, r, and r+d, respectively, where r is the middle root and d is the common difference between the roots.

Step 2: Apply Vieta's formulas

Vieta's formulas relate the coefficients of a polynomial to the sums and products of its roots. For a cubic equation \(ax^3 + bx^2 + cx + d = 0\), the formulas are: 1. Sum of roots: \((r-d) + r + (r+d) = (\frac{-b}{a})\) 2. Sum of pairwise products of roots: \((r-d)r + (r-d)(r+d) + r(r+d) = (\frac{c}{a})\) 3. Product of roots: \((r-d)r(r+d) = (\frac{-d}{a})\) We will apply these formulas to our cubic equation \(4x^3 - 24x^2 + 23x +18 = 0\).

Step 3: Use the sum of roots

The sum of the roots is given by: \((r-d) + r + (r+d) = \frac{-24}{4}\) Simplifying and solving for r: \(3r = -6\) \(r = -2\) So, the middle root r is -2.

Step 4: Find the sum of pairwise products of roots

Now we'll find the sum of pairwise products of the roots: \((r-d)r + (r-d)(r+d) + r(r+d) = \frac{23}{4}\) By substituting the value of r that we found in step 3: \((-2-d)(-2) + (-2-d)(-2+d) + (-2)(-2+d) = \frac{23}{4}\) Solving, we simplily this equation as: \(2d^2 + 6d - 23 = 0\)

Step 5: Solve the quadratic equation for d

Now we need to solve for d in the quadratic equation: \(2d^2 + 6d - 23 = 0\) Applying the quadratic formula: \(d = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-6 \pm \sqrt{6^2 - 4(2)(-23)}}{2(2)}\) After calculation, we find out that there are two possible values for d: \(d = 2\) and \(d = -\frac{23}{2}\). However, since the roots are in arithmetic progression, the common difference must be rational. Thus, we choose \(d = 2\).

Step 6: Find the roots

Now that we have the values for r and d, we can find all the three roots: 1. First root: \(r-d = -2 - 2 = -4\) 2. Second root (middle): \(r = -2\) 3. Third root: \(r+d = -2 + 2 = 0\) So the roots of the given equation are -4, -2, and 0.

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Chapter 21: Chapter 21

Solve the equation \(4 \mathrm{x}^{3}-24 \mathrm{x}^{2}+23 \mathrm{x}+18=0\), having given that the roots are in arithmetical progression.

Short Answer

Step by step solution

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Step 1: Set up the roots

Step 2: Apply Vieta's formulas

Step 3: Use the sum of roots

Step 4: Find the sum of pairwise products of roots

Step 5: Solve the quadratic equation for d

Step 6: Find the roots

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