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Chapter 19: Problem 1804

If the roots of the quadratic equation \(x^{2}+A x+B=0\) are \(\tan 30^{\circ}\) and \(\tan 15^{\circ}\) then the value of \(A-B=\) (a) 1 (b) \(-1\) (c) 2 (d) 3

Short Answer

Expert verified
The value of \(A - B = 1\). Therefore, the correct answer is (a).

Step by step solution

01

Find the sum and product of the roots

Since we know the roots, we can directly find their sum and product: Sum of roots = \(\tan 30^{\circ} + \tan 15^{\circ}\) Product of roots = \(\tan 30^{\circ} \cdot \tan 15^{\circ}\)
02

Use Vieta's formulas to relate the sum and product of roots to A and B

According to Vieta's formulas, we have: \(A = -(sum \; of \; roots),\) therefore, \(A = -\tan 30^{\circ} - \tan 15^{\circ}\) \(B = product \; of \; roots,\) therefore, \(B = \tan 30^{\circ} \cdot \tan 15^{\circ}\)
03

Find A - B

Now, we need to find \(A - B\): \(A - B = (-\tan 30^{\circ} - \tan 15^{\circ}) - (\tan 30^{\circ} \cdot \tan 15^{\circ})\) Since \(\tan 30^{\circ} = \frac{1}{\sqrt{3}}\) and \(\tan 15^{\circ} = 2 - \sqrt{3}\), we can plug in these values: \(A - B = (-\frac{1}{\sqrt{3}} - (2 - \sqrt{3})) - (\frac{1}{\sqrt{3}} \cdot (2 - \sqrt{3}))\)
04

Calculate A - B

Simplify the expression for \(A - B\): \(A - B = -\frac{1}{\sqrt{3}} - 2 + \sqrt{3} - (\frac{2}{\sqrt{3}} - 1)\) Now, combine like terms and simplify further: \(A - B = -\frac{1}{\sqrt{3}} + \sqrt{3} + 1 - \frac{2}{\sqrt{3}}\) \(A - B = (\frac{- 1 - 2}{\sqrt{3}}) + \sqrt{3} + 1\) \(A - B = -\frac{3}{\sqrt{3}} + \sqrt{3} + 1\) Now, rationalize the denominator: \(A - B = -\sqrt{3} + \sqrt{3} + 1\) \(A - B = 1\) So, the value of \(A - B = 1\), and the answer is (a).

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