Question

If the roots of the quadratic equation \(ax^2 + bx+c=0\) are \(sin α\) and \(cos α\), then \(1+2\frac{c}{a}=\)

• A. \(\frac{a^2}{b^2}\)
• B. \(\frac{b^2}{a^2}\)
• C. \(a^2\)
• D. \(b^2\)

Answer 1

The Correct Option is B

To solve the problem, we need to determine the value of \(1 + 2\frac{c}{a}\) given that the roots of the quadratic equation \(ax^2 + bx + c = 0\) are \(\sin \alpha\) and \(\cos \alpha\).

1. Using Vieta's Formulas:
For a quadratic equation \(ax^2 + bx + c = 0\) with roots \(\sin \alpha\) and \(\cos \alpha\), Vieta's formulas tell us: \[ \sin \alpha + \cos \alpha = -\frac{b}{a} \] \[ \sin \alpha \cdot \cos \alpha = \frac{c}{a} \]

2. Expressing \(\sin \alpha + \cos \alpha\) and \(\sin \alpha \cdot \cos \alpha\):
We know from trigonometric identities that: \[ (\sin \alpha + \cos \alpha)^2 = \sin^2 \alpha + \cos^2 \alpha + 2 \sin \alpha \cos \alpha \] Since \(\sin^2 \alpha + \cos^2 \alpha = 1\), we can rewrite this as: \[ (\sin \alpha + \cos \alpha)^2 = 1 + 2 \sin \alpha \cos \alpha \] Substituting \(\sin \alpha + \cos \alpha = -\frac{b}{a}\) and \(\sin \alpha \cdot \cos \alpha = \frac{c}{a}\), we get: \[ \left(-\frac{b}{a}\right)^2 = 1 + 2 \cdot \frac{c}{a} \] \[ \frac{b^2}{a^2} = 1 + 2 \cdot \frac{c}{a} \]

3. Solving for \(1 + 2 \frac{c}{a}\):
From the equation above, we directly see that: \[ 1 + 2 \cdot \frac{c}{a} = \frac{b^2}{a^2} \]

Final Answer:
The value of \(1 + 2 \frac{c}{a}\) is \({\frac{b^2}{a^2}}\).