By Vieta's formulas, we know that $\alpha + \beta = \frac{5}{2}$ and $\alpha\beta = \frac{m}{2}$. Therefore, $\alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2\alpha\beta = \frac{25}{4} - m$.
Now, we want to find $\alpha^2+\beta^2$ in terms of $m$ in order to find an expression for $\alpha^2+\beta^2$.
Taking the square of $\alpha + \beta$, we can write:
$$(\alpha+\beta)^2 = \alpha^2 + 2\alpha\beta + \beta^2 = \frac{25}{4}$$
Substituting $\alpha\beta = \frac{m}{2}$, we get:
$$\alpha^2+\beta^2 = \left(\frac{5}{2}\right)^2 - 2\cdot\frac{m}{2} = \frac{25-2m}{4}$$
Therefore, $\alpha^2+\beta^2 = \frac{25-2m}{4}$, so $\alpha^2+\beta^2+2\alpha\beta = \frac{25-2m}{4} + 2\cdot\frac{m}{2}$.
Simplifying, we get:
$$\alpha^2+\beta^2+2\alpha\beta = \frac{25+2m}{4}$$
Now, we can express $\alpha^2 + \beta^2$ in terms of $\alpha^2 + \beta^2 + 2\alpha\beta$ by subtracting $2\alpha\beta$ from both sides:
$$\alpha^2+\beta^2 = \frac{25+2m}{4} - 2\alpha\beta$$
Substituting $\alpha\beta = \frac{m}{2}$:
$$\alpha^2+\beta^2 = \frac{25+2m}{4} - m = \frac{25-2m}{4}$$
Therefore, $\alpha^2+\beta^2 = \frac{25-2m}{4}$, so the answer is $\boxed{\textbf{(C) } 25-2m}$.
The roots of the quadratic equation 2𝑥2 − 5𝑥 +
𝑚 = 0 are ∝ 𝑎𝑛𝑑 𝛽, where 𝑚 is a real number.
Find an expression for ∝2+ 𝛽2
in terms of 𝑚.
A. 25 + 2𝑚
B. 25
4 − 𝑚
C. 25 −2m
D. 25
1 answer