If the roots of the quadratic equation x^2 + px + q = 0 are \tan 30^ \circ and \tan 15^ \circ , respectively, then the value of 2 + q - p is

Quadratic Equation and Inequalities PYQ — JEE Main 2006 · Shift 0 · Q69 with Solution

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Correct answer: B

Let the roots of $x^2+px+q=0$ be $\alpha=\tan 30^\circ,\qquad \beta=\tan 15^\circ.$
For a quadratic equation $x^2+px+q=0$ , by Vieta’s formulas: $\alpha+\beta=-p,\qquad \alpha\beta=q.$
Compute the roots: $\tan 30^\circ=\frac{1}{\sqrt{3}}.$ Also,

=\frac{\tan45^\circ-\tan30^\circ}{1+\tan45^\circ\tan30^\circ} =\frac{1-\frac{1}{\sqrt3}}{1+\frac{1}{\sqrt3}}.$$ Multiply numerator and denominator by $\sqrt3$: $$\tan15^\circ=\frac{\sqrt3-1}{\sqrt3+1} =\frac{(\sqrt3-1)^2}{3-1} =\frac{3+1-2\sqrt3}{2}=2-\sqrt3.$$ So, $$\alpha=\frac{1}{\sqrt3},\qquad \beta=2-\sqrt3.$$ 4. Find the sum: $$\alpha+\beta=\frac{1}{\sqrt3}+2-\sqrt3.$$ Since $\frac{1}{\sqrt3}=\frac{\sqrt3}{3}$, $$\alpha+\beta=2+\frac{\sqrt3}{3}-\sqrt3=2-\frac{2\sqrt3}{3}.$$ Hence, $$p=-(\alpha+\beta)=-2+\frac{2\sqrt3}{3}.$$ 5. Find the product: $$q=\alpha\beta=\frac{1}{\sqrt3}(2-\sqrt3)=\frac{2}{\sqrt3}-1=\frac{2\sqrt3}{3}-1.$$ 6. Now compute: $$2+q-p=2+\left(\frac{2\sqrt3}{3}-1\right)-\left(-2+\frac{2\sqrt3}{3}\right).$$ Simplify: $$2+q-p=2+\frac{2\sqrt3}{3}-1+2-\frac{2\sqrt3}{3}=3.$$ 7. Therefore, the correct option is $$\boxed{3}$$ which is option **B**.

View written solutionFree

Correct answer: B

Let the roots of $x^2+px+q=0$ be $\alpha=\tan 30^\circ,\qquad \beta=\tan 15^\circ.$
For a quadratic equation $x^2+px+q=0$ , by Vieta’s formulas: $\alpha+\beta=-p,\qquad \alpha\beta=q.$
Compute the roots: $\tan 30^\circ=\frac{1}{\sqrt{3}}.$ Also,

=\frac{\tan45^\circ-\tan30^\circ}{1+\tan45^\circ\tan30^\circ} =\frac{1-\frac{1}{\sqrt3}}{1+\frac{1}{\sqrt3}}.$$ Multiply numerator and denominator by $\sqrt3$: $$\tan15^\circ=\frac{\sqrt3-1}{\sqrt3+1} =\frac{(\sqrt3-1)^2}{3-1} =\frac{3+1-2\sqrt3}{2}=2-\sqrt3.$$ So, $$\alpha=\frac{1}{\sqrt3},\qquad \beta=2-\sqrt3.$$ 4. Find the sum: $$\alpha+\beta=\frac{1}{\sqrt3}+2-\sqrt3.$$ Since $\frac{1}{\sqrt3}=\frac{\sqrt3}{3}$, $$\alpha+\beta=2+\frac{\sqrt3}{3}-\sqrt3=2-\frac{2\sqrt3}{3}.$$ Hence, $$p=-(\alpha+\beta)=-2+\frac{2\sqrt3}{3}.$$ 5. Find the product: $$q=\alpha\beta=\frac{1}{\sqrt3}(2-\sqrt3)=\frac{2}{\sqrt3}-1=\frac{2\sqrt3}{3}-1.$$ 6. Now compute: $$2+q-p=2+\left(\frac{2\sqrt3}{3}-1\right)-\left(-2+\frac{2\sqrt3}{3}\right).$$ Simplify: $$2+q-p=2+\frac{2\sqrt3}{3}-1+2-\frac{2\sqrt3}{3}=3.$$ 7. Therefore, the correct option is $$\boxed{3}$$ which is option **B**.