Question

If $4+3x-7x^2$ attains its maximum value $M$ at $x=\alpha$ and $5x^2-2x+1$ attains its minimum value at $x=\beta$, then $$\frac{28\,(M - \alpha)}{5\,(m + \beta)} =\,?$$ \textit{(Assume $m$ is that minimum value of $5x^2 -2x +1$ at $x=\beta$)}

• A. $28$
• B. $23$
• C. $5$
• D. $1$

Hint:

For a quadratic $ax^2+bx+c$, the vertex $x$-coordinate is $-\tfrac{b}{2a}$.
- Always substitute back carefully to find the extremum (maximum or minimum) value.

Answer 1

The Correct Option is B

Step 1: Eliminate the radicals and rearrange.

Given the equation:

$$\frac{x - 1}{\sqrt{2x^2 - 5x + 2}} = \frac{41}{60},$$

cross-multiply to obtain:

$$60(x - 1) = 41\sqrt{2x^2 - 5x + 2}.$$

Next, square both sides (being careful) and move all terms to one side to form a polynomial equation in $x$.

Step 2: Solve the resulting equation.

Expanding both sides, we get:

$$3600(x - 1)^2 = 1681(2x^2 - 5x + 2).$$

Simplify this expression and solve for $x$. This process should yield two real solutions, though there may be extraneous solutions to check.

Step 3: Select the root in the interval $-\tfrac{1}{2} < x < 0$.

Among the real solutions, determine which one falls between $-\tfrac{1}{2}$ and $0$. The correct root is $\boxed{-\tfrac{7}{34}}$.