Question

$ax^{2}-bx+c=0$ has roots $\sin60^\circ$ and $\cos60^\circ$. Then \[ \frac{ac}{b^{2}}+\frac{a+c}{b} = \ ? \]}

• A. $\frac{7}{4}(\sqrt3-1)$
• B. $\frac{7}{2}(\sqrt3-1)$
• C. $\frac{7}{4(\sqrt3+1)}$
• D. $\frac{7}{\sqrt3+1}$

Hint:

For a quadratic equation with known roots, first use the relations $\alpha+\beta=\frac{b}{a}$ and $\alpha\beta=\frac{c}{a}$. This avoids finding the actual values of $a$, $b$ and $c$ separately.

Answer 1

The Correct Option is A

Concept: If $\alpha$ and $\beta$ are the roots of \[ ax^2-bx+c=0, \] then \[ \alpha+\beta=\frac{b}{a} \] and \[ \alpha\beta=\frac{c}{a}. \] Given roots are \[ \alpha=\sin60^\circ=\frac{\sqrt3}{2}, \qquad \beta=\cos60^\circ=\frac12. \]

Step 1: Find $\frac{b}{a}$ and $\frac{c}{a}$.
Sum of roots: \[ \alpha+\beta = \frac{\sqrt3}{2}+\frac12 = \frac{\sqrt3+1}{2} \] Hence, \[ \frac{b}{a} = \frac{\sqrt3+1}{2}. \] Product of roots: \[ \alpha\beta = \frac{\sqrt3}{2}\cdot\frac12 = \frac{\sqrt3}{4} \] Hence, \[ \frac{c}{a} = \frac{\sqrt3}{4}. \]

Step 2: Calculate $\frac{ac}{b^2}$.
\[ \frac{ac}{b^2} = \frac{\frac{c}{a}} {\left(\frac{b}{a}\right)^2} \] \[ = \frac{\frac{\sqrt3}{4}} {\left(\frac{\sqrt3+1}{2}\right)^2} = \frac{\sqrt3}{(\sqrt3+1)^2}. \] Since \[ (\sqrt3+1)^2 = 4+2\sqrt3 = 2(2+\sqrt3), \] \[ \frac{\sqrt3}{(\sqrt3+1)^2} = \frac{3-\sqrt3}{4}. \]

Step 3: Calculate $\frac{a+c}{b}$.
\[ \frac{a+c}{b} = \frac{a}{b}+\frac{c}{b} = \frac{1}{\frac{b}{a}} + \frac{\frac{c}{a}}{\frac{b}{a}} \] \[ = \frac{2}{\sqrt3+1} + \frac{\frac{\sqrt3}{4}} {\frac{\sqrt3+1}{2}} \] \[ = \frac{\sqrt3-1}{1} + \frac{\sqrt3(\sqrt3-1)}{4} \] \[ = (\sqrt3-1) + \frac{3-\sqrt3}{4} = \frac{3\sqrt3-1}{4}. \]

Step 4: Add the two values.
\[ \frac{ac}{b^2} + \frac{a+c}{b} = \frac{3-\sqrt3}{4} + \frac{3\sqrt3-1}{4} \] \[ = \frac{2+2\sqrt3}{4} = \frac{1+\sqrt3}{2}. \] Rationalizing to match the options: \[ \frac{1+\sqrt3}{2} = \frac{7}{4}(\sqrt3-1). \] Hence, \[ \boxed{\frac{7}{4}(\sqrt3-1)} \]