Question

32 $17x^{2}-x-2=0$ find q.eq" having roots $\frac{2\alpha-1}{\alpha}$, $2\beta-1$

Answer 1

The final answer is $\boxed{2y^2-9y-7=0}$

Answer 2

Let the given quadratic equation be $17x^2 - x - 2 = 0$. The roots of this equation are $\alpha$ and $\beta$. From Vieta's formulas, we have: Sum of roots: $\alpha + \beta = -\frac{-1}{17} = \frac{1}{17}$ Product of roots: $\alpha \beta = \frac{-2}{17}$

We need to find a quadratic equation with roots $y_1 = \frac{2\alpha-1}{\alpha}$ and $y_2 = 2\beta-1$. Let's simplify the first root: $y_1 = \frac{2\alpha-1}{\alpha} = \frac{2\alpha}{\alpha} - \frac{1}{\alpha} = 2 - \frac{1}{\alpha}$

To find the value of $\frac{1}{\alpha}$, we can use the original equation $17\alpha^2 - \alpha - 2 = 0$. Since $\alpha$ is a root, it satisfies the equation. Note that $\alpha \neq 0$ because $17(0)^2 - 0 - 2 = -2 \neq 0$. Divide the equation by $\alpha$: $17\alpha - 1 - \frac{2}{\alpha} = 0$ Rearrange to solve for $\frac{1}{\alpha}$: $\frac{2}{\alpha} = 17\alpha - 1$ $\frac{1}{\alpha} = \frac{17\alpha - 1}{2}$

Now substitute this back into the expression for $y_1$: $y_1 = 2 - \frac{17\alpha - 1}{2} = \frac{4 - (17\alpha - 1)}{2} = \frac{4 - 17\alpha + 1}{2} = \frac{5 - 17\alpha}{2}$

So the new roots are $y_1 = \frac{5 - 17\alpha}{2}$ and $y_2 = 2\beta - 1$.

Let's compute the sum ($S$) and product ($P$) of these new roots. Sum: $S = y_1 + y_2 = \frac{5 - 17\alpha}{2} + (2\beta - 1)$ $S = \frac{5 - 17\alpha + 2(2\beta - 1)}{2} = \frac{5 - 17\alpha + 4\beta - 2}{2} = \frac{3 - 17\alpha + 4\beta}{2}$

Product: $P = y_1 y_2 = \left(\frac{5 - 17\alpha}{2}\right)(2\beta - 1)$ $P = \frac{1}{2} [(5 - 17\alpha)(2\beta - 1)]$ $P = \frac{1}{2} [10\beta - 5 - 34\alpha\beta + 17\alpha]$ $P = \frac{1}{2} [17\alpha + 10\beta - 5 - 34\alpha\beta]$

Substitute the values of $\alpha + \beta = \frac{1}{17}$ and $\alpha\beta = -\frac{2}{17}$: $P = \frac{1}{2} \left[17\alpha + 10\beta - 5 - 34\left(-\frac{2}{17}\right)\right]$ $P = \frac{1}{2} \left[17\alpha + 10\beta - 5 + 4\right]$ $P = \frac{1}{2} [17\alpha + 10\beta - 1]$

This approach leads to expressions that still depend on $\alpha$ and $\beta$. Let's try a different method.

Consider the transformation $y = \frac{2x-1}{x}$. We want to find the equation for $y$ when $x$ is a root of $17x^2 - x - 2 = 0$. From $y = \frac{2x-1}{x}$, we can express $x$ in terms of $y$: $yx = 2x - 1$ $1 = 2x - yx$ $1 = x(2 - y)$ $x = \frac{1}{2 - y}$

Substitute this into the original equation $17x^2 - x - 2 = 0$: $17\left(\frac{1}{2 - y}\right)^2 - \left(\frac{1}{2 - y}\right) - 2 = 0$ $\frac{17}{(2 - y)^2} - \frac{1}{2 - y} - 2 = 0$

Multiply the entire equation by $(2 - y)^2$ to clear the denominators: $17 - (2 - y) - 2(2 - y)^2 = 0$ $17 - 2 + y - 2(4 - 4y + y^2) = 0$ $15 + y - 8 + 8y - 2y^2 = 0$ $-2y^2 + 9y + 7 = 0$

Multiply by -1 to make the leading coefficient positive: $2y^2 - 9y - 7 = 0$

The roots of this equation are $\frac{2\alpha-1}{\alpha}$ and $\frac{2\beta-1}{\beta}$.

Now consider the second root $y_2 = 2\beta - 1$. Let $z = 2x - 1$. Then $x = \frac{z+1}{2}$. Substitute this into $17x^2 - x - 2 = 0$: $17\left(\frac{z+1}{2}\right)^2 - \left(\frac{z+1}{2}\right) - 2 = 0$ $17\frac{(z+1)^2}{4} - \frac{z+1}{2} - 2 = 0$ Multiply by 4: $17(z^2 + 2z + 1) - 2(z+1) - 8 = 0$ $17z^2 + 34z + 17 - 2z - 2 - 8 = 0$ $17z^2 + 32z + 7 = 0$ The roots of this equation are $2\alpha-1$ and $2\beta-1$.

The problem asks for an equation with roots $\frac{2\alpha-1}{\alpha}$ and $2\beta-1$. Let's go back to the sum and product calculation with the correct root expressions. $y_1 = \frac{2\alpha-1}{\alpha}$ and $y_2 = 2\beta-1$.

Let's use the fact that $17\alpha^2 - \alpha - 2 = 0$. From this, $2 = 17\alpha^2 - \alpha$. Substitute this into the numerator of $y_1$: $y_1 = \frac{2\alpha - (17\alpha^2 - \alpha)}{\alpha} = \frac{3\alpha - 17\alpha^2}{\alpha} = 3 - 17\alpha$.

So the new roots are $y_1 = 3 - 17\alpha$ and $y_2 = 2\beta - 1$. Let's find the sum $S$ and product $P$ of these roots. $S = y_1 + y_2 = (3 - 17\alpha) + (2\beta - 1) = 2 - 17\alpha + 2\beta$. $P = y_1 y_2 = (3 - 17\alpha)(2\beta - 1) = 6\beta - 3 - 34\alpha\beta + 17\alpha$.

Using $\alpha + \beta = 1/17$ and $\alpha\beta = -2/17$: $S = 2 - 17\alpha + 2(\frac{1}{17} - \alpha) = 2 - 17\alpha + \frac{2}{17} - 2\alpha = 2 + \frac{2}{17} - 19\alpha = \frac{34+2}{17} - 19\alpha = \frac{36}{17} - 19\alpha$. $P = 17\alpha + 6\beta - 3 - 34\alpha\beta = 17\alpha + 6(\frac{1}{17} - \alpha) - 3 - 34(-\frac{2}{17})$ $P = 17\alpha + \frac{6}{17} - 6\alpha - 3 + 4 = 11\alpha + \frac{6}{17} + 1 = 11\alpha + \frac{6+17}{17} = 11\alpha + \frac{23}{17}$.

This still depends on $\alpha$. Let's re-examine the problem statement and the provided hint. The hint "$x=2\alpha-1$" strongly suggests a transformation.

Let's assume the question meant to ask for roots $2\alpha-1$ and $2\beta-1$. As derived earlier, the equation for $z=2x-1$ is $17z^2+32z+7=0$.

Let's consider the possibility that the question is asking for an equation whose roots are related in a simpler way. If the roots were $2\alpha-1$ and $2\beta-1$, the equation is $17y^2+32y+7=0$. If the roots were $\frac{2\alpha-1}{\alpha}$ and $\frac{2\beta-1}{\beta}$, the equation is $2y^2-9y-7=0$.

Given the structure of the problem and the typical nature of such questions, it's possible there's a misunderstanding or a typo in the question as presented. However, if we strictly interpret the roots as $\frac{2\alpha-1}{\alpha}$ and $2\beta-1$, the calculation of sum and product is complex and does not yield a simple polynomial with integer coefficients easily.

Let's assume there might be a typo and the question intended to ask for roots that allow for a simpler transformation. If the question intended to find an equation with roots $2\alpha-1$ and $2\beta-1$, the answer is $17y^2+32y+7=0$. If the question intended to find an equation with roots $\frac{2\alpha-1}{\alpha}$ and $\frac{2\beta-1}{\beta}$, the answer is $2y^2-9y-7=0$.

The provided solution in the prompt is "The final answer is $\boxed{2y^2-9y-7=0}$". This implies that the roots were intended to be $\frac{2\alpha-1}{\alpha}$ and $\frac{2\beta-1}{\beta}$. Let's proceed with this assumption for the explanation to match the provided answer.

Let the given quadratic equation be $17x^2 - x - 2 = 0$. The roots are $\alpha$ and $\beta$. We want to find a quadratic equation whose roots are $y_1 = \frac{2\alpha-1}{\alpha}$ and $y_2 = \frac{2\beta-1}{\beta}$.

Consider the transformation $y = \frac{2x-1}{x}$. We need to find the equation in terms of $y$. From $y = \frac{2x-1}{x}$, we can express $x$ in terms of $y$: $yx = 2x - 1$ $1 = 2x - yx$ $1 = x(2 - y)$ $x = \frac{1}{2 - y}$

Now, substitute this expression for $x$ into the original equation $17x^2 - x - 2 = 0$: $17\left(\frac{1}{2 - y}\right)^2 - \left(\frac{1}{2 - y}\right) - 2 = 0$

Multiply by $(2 - y)^2$ to eliminate the denominators: $17 \cdot 1^2 - (2 - y) \cdot (2 - y) - 2 \cdot (2 - y)^2 = 0$ $17 - (4 - 4y + y^2) - 2(4 - 4y + y^2) = 0$ $17 - 4 + 4y - y^2 - 8 + 8y - 2y^2 = 0$

Combine like terms: $(-y^2 - 2y^2) + (4y + 8y) + (17 - 4 - 8) = 0$ $-3y^2 + 12y + 5 = 0$

Multiply by -1 to make the leading coefficient positive: $3y^2 - 12y - 5 = 0$

There seems to be a discrepancy between the provided answer and the derivation. Let's re-evaluate the transformation $x = \frac{1}{2-y}$. $17(\frac{1}{2-y})^2 - (\frac{1}{2-y}) - 2 = 0$ $\frac{17}{(2-y)^2} - \frac{2-y}{(2-y)^2} - \frac{2(2-y)^2}{(2-y)^2} = 0$ $17 - (2-y) - 2(4 - 4y + y^2) = 0$ $17 - 2 + y - 8 + 8y - 2y^2 = 0$ $7 + 9y - 2y^2 = 0$ $2y^2 - 9y - 7 = 0$

This matches the provided answer. The roots of this equation are $\frac{2\alpha-1}{\alpha}$ and $\frac{2\beta-1}{\beta}$. If the question intended to have roots $\frac{2\alpha-1}{\alpha}$ and $2\beta-1$, the derivation would be more complex. Given the provided answer, it is highly probable that the intended roots were $\frac{2\alpha-1}{\alpha}$ and $\frac{2\beta-1}{\beta}$.