Question

Let \(\tan 30^\circ\) and \(\tan 15^\circ\) be the roots of the quadratic equation \[ x^2+ax+b=0, \] then \(1+a-b=\)

• A. \(0\)
• B. \(1\)
• C. \(ab\)
• D. \(a^2b^2\)

Hint:

For a quadratic equation \(x^2+ax+b=0\), remember that sum of roots is \(-a\) and product of roots is \(b\).

Answer 1

The Correct Option is A

Step 1: Use sum and product of roots.
For the quadratic equation \[ x^2+ax+b=0, \] if roots are \(\alpha\) and \(\beta\), then \[ \alpha+\beta=-a \] and \[ \alpha\beta=b. \] Here, \[ \alpha=\tan30^\circ,\qquad \beta=\tan15^\circ. \]

Step 2: Find the values of the roots.
We know that \[ \tan30^\circ=\frac{1}{\sqrt{3}}. \] Also, \[ \tan15^\circ=\tan(45^\circ-30^\circ). \] Using \[ \tan(A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}, \] we get \[ \tan15^\circ = \frac{1-\frac{1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}}}. \] \[ = \frac{\sqrt{3}-1}{\sqrt{3}+1}. \] Rationalizing, \[ \tan15^\circ=2-\sqrt{3}. \]

Step 3: Find \(a\) and \(b\).
Now, \[ \alpha+\beta = \frac{1}{\sqrt{3}}+2-\sqrt{3}. \] Since \[ \alpha+\beta=-a, \] we get \[ a=-\left(\frac{1}{\sqrt{3}}+2-\sqrt{3}\right). \] Also, \[ b=\alpha\beta = \frac{1}{\sqrt{3}}(2-\sqrt{3}). \] \[ b=\frac{2-\sqrt{3}}{\sqrt{3}}. \]

Step 4: Evaluate \(1+a-b\).
\[ 1+a-b = 1-\left(\frac{1}{\sqrt{3}}+2-\sqrt{3}\right)-\frac{2-\sqrt{3}}{\sqrt{3}}. \] \[ = 1-\frac{1}{\sqrt{3}}-2+\sqrt{3}-\frac{2}{\sqrt{3}}+1. \] \[ = \sqrt{3}-\frac{3}{\sqrt{3}}. \] Since \[ \frac{3}{\sqrt{3}}=\sqrt{3}, \] we get \[ 1+a-b=0. \]

Step 5: Final conclusion.
Therefore, \[ \boxed{0} \]