JEE Quadratic 6

It is not too hard to use the 'half angle formulas' to show that

$\tan{\dfrac{3\pi}{8}}=1+\sqrt{2}$ and so a root of the polynomial is

$\tan^{2}\dfrac{3\pi}{8}=(1+\sqrt{2})^{2}=3+2\sqrt{2}$

Now irrational roots of a quadratic occur in pairs (in a similar way to complex roots), and so the other root of the quadratic is $3-2\sqrt{2}$

So we can write the quadratic equation as

$(x-3-2\sqrt{2})(x-3+2\sqrt{2})=0$

$\implies x^{2}-6x+1=0$

$\implies 2x^2-3\times 4x+2=0$

Comparing this with the form of the equation in the question gives

$a=4,b=2 \implies a+b=\boxed{6}$

Tan^2(3pie/8)=3+2{(2)^1/2}=y(let) x+3+root8=rational(3a/2) x=R-root8 thus x must be rational +(-root8) let root8=z,rational no=k x=k-z Tan^2(3pie/8)=3+z x Tan^2(3pie/8)=b/2(rational)=(k-z)(3+z)=3k+(z k)-(3 z)-z z irrational part must be made 0=k z-3 z=z(k-3) k=3 hence other root is =3-2{(2)^1/2} --->a=4,b=2 hence a+b=6

We have the equality $\tan\left(\frac{\theta}2\right)=\frac{1-\cos\theta}{\sin\theta}$ , which gives:

$\tan^2\left(\frac{3\pi}8\right) = \frac{(1-\cos(\frac{3\pi}4))^2}{\sin^2(\frac{3\pi}4)} =3+2\sqrt2.$

The roots of the polynomial are:

$\frac{3a\pm\sqrt{9a^2-32b}}4$

giving us $a=4$ and $b=\frac12$ , hence $a+4b=\boxed{6}$ .