Question

If $\alpha$ and $\beta$ are the roots of the equation 2x (2x + 1) = 1, then $\beta$ is equal to

• A. $2\alpha^2$
• B. $-2\alpha (\alpha + 1)$
• C. $2\alpha (\alpha - 1)$
• D. $2\alpha (\alpha + 1)$

Answer 1

Answer: (B)

(B)

Answer 2

The given equation is $2x(2x + 1) = 1$. Expanding this, we get $4x^2 + 2x = 1$. Rearranging into the standard quadratic form $ax^2 + bx + c = 0$, we have $4x^2 + 2x - 1 = 0$.

Let $\alpha$ and $\beta$ be the roots of this equation. According to Vieta's formulas, the sum of the roots is $\alpha + \beta = -\frac{b}{a}$ and the product of the roots is $\alpha \beta = \frac{c}{a}$. For the equation $4x^2 + 2x - 1 = 0$, we have $a=4$, $b=2$, and $c=-1$. The sum of the roots is $\alpha + \beta = -\frac{2}{4} = -\frac{1}{2}$. The product of the roots is $\alpha \beta = \frac{-1}{4} = -\frac{1}{4}$.

From the sum of roots, we can express $\beta$ in terms of $\alpha$: $\beta = -\frac{1}{2} - \alpha$.

Since $\alpha$ is a root of the equation $4x^2 + 2x - 1 = 0$, it must satisfy the equation: $4\alpha^2 + 2\alpha - 1 = 0$. From this equation, we can express $4\alpha^2$ or $2\alpha^2$ in terms of $\alpha$: $4\alpha^2 = 1 - 2\alpha$ and $2\alpha^2 = \frac{1 - 2\alpha}{2} = \frac{1}{2} - \alpha$.

Now let's evaluate the given options using the relation $2\alpha^2 = \frac{1}{2} - \alpha$ and compare them with the expression for $\beta$, which is $-\frac{1}{2} - \alpha$.

(A) $2\alpha^2 = \frac{1}{2} - \alpha$. This is not equal to $-\frac{1}{2} - \alpha$. (B) $-2\alpha (\alpha + 1) = -2\alpha^2 - 2\alpha$. Substitute $2\alpha^2 = \frac{1}{2} - \alpha$: $-2\alpha^2 - 2\alpha = -(\frac{1}{2} - \alpha) - 2\alpha = -\frac{1}{2} + \alpha - 2\alpha = -\frac{1}{2} - \alpha$. This matches the expression for $\beta$. (C) $2\alpha (\alpha - 1) = 2\alpha^2 - 2\alpha$. Substitute $2\alpha^2 = \frac{1}{2} - \alpha$: $2\alpha^2 - 2\alpha = (\frac{1}{2} - \alpha) - 2\alpha = \frac{1}{2} - 3\alpha$. This is not equal to $-\frac{1}{2} - \alpha$. (D) $2\alpha (\alpha + 1) = 2\alpha^2 + 2\alpha$. Substitute $2\alpha^2 = \frac{1}{2} - \alpha$: $2\alpha^2 + 2\alpha = (\frac{1}{2} - \alpha) + 2\alpha = \frac{1}{2} + \alpha$. This is not equal to $-\frac{1}{2} - \alpha$.

Thus, option (B) is the correct expression for $\beta$ in terms of $\alpha$.