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Problem 631

# 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.

Expert verified
The roots of the given equation $$4x^3 - 24x^2 + 23x +18 = 0$$ are -4, -2, and 0.
See the step by step solution

## 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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