If \(\alpha~,\beta\) are the roots of the equation \(x^2 - px+q=0~\text{and}~\alpha\) > \(0, \beta\) >\(0\) , then the value of \(\alpha^{\frac{1}{4}}+\beta^{\frac{1}{4}}~\text{is}~\left ( p+6 \sqrt{q}+4q^{\frac{1}{4}} \sqrt{p+2\sqrt{q}} \right )^k\) , where k is equal to
If \(\alpha~,\beta\) are the roots of the equation \(x^2 - px+q=0~\text{and}~\alpha\) > \(0, \beta\) >\(0\) , then the value of \(\alpha^{\frac{1}{4}}+\beta^{\frac{1}{4}}~\text{is}~\left ( p+6 \sqrt{q}+4q^{\frac{1}{4}} \sqrt{p+2\sqrt{q}} \right )^k\) , where k is equal to
\(x^2-px+ q=0 \\ \therefore \alpha + \beta = p~\text{and}~\alpha \beta = q. \)
\(\text{Now}, ~\left ( \alpha^{\frac{1}{4}}+\beta^{\frac{1}{4}} \right )^4=\left [ \left ( \alpha^{\frac{1}{4}}+\beta^{\frac{1}{4}} \right )^2 \right ]^2 \\ =\left [ \alpha^{\frac{1}{2}}+\beta^{\frac{1}{2}}+2(\alpha \beta)^{\frac{1}{4}} \right ]^2\)
\(=\left [ \sqrt{\alpha+\beta+2\sqrt{\alpha \beta}}+2(\alpha \beta)^{\frac{1}{4}} \right ]^2 \\ =\left [ \sqrt{p+~2\sqrt{q}}+2(q)^{\frac{1}{4}} \right ]^2 \\ =p+6 \sqrt{q}+4q^{\frac{1}{4}}\sqrt{p+2\sqrt{q}}\\ \therefore \alpha^{\frac{1}{4}}+\beta^{\frac{1}{4}}=\left [ p+6\sqrt{q}+4q^{\frac{1}{4}}\sqrt{p+2\sqrt{q}} \right ]^{\frac{1}{4}}\)
\(x^2-px+ q=0 \\ \therefore \alpha + \beta = p~\text{and}~\alpha \beta = q. \)
\(\text{Now}, ~\left ( \alpha^{\frac{1}{4}}+\beta^{\frac{1}{4}} \right )^4=\left [ \left ( \alpha^{\frac{1}{4}}+\beta^{\frac{1}{4}} \right )^2 \right ]^2 \\ =\left [ \alpha^{\frac{1}{2}}+\beta^{\frac{1}{2}}+2(\alpha \beta)^{\frac{1}{4}} \right ]^2\)
\(=\left [ \sqrt{\alpha+\beta+2\sqrt{\alpha \beta}}+2(\alpha \beta)^{\frac{1}{4}} \right ]^2 \\ =\left [ \sqrt{p+~2\sqrt{q}}+2(q)^{\frac{1}{4}} \right ]^2 \\ =p+6 \sqrt{q}+4q^{\frac{1}{4}}\sqrt{p+2\sqrt{q}}\\ \therefore \alpha^{\frac{1}{4}}+\beta^{\frac{1}{4}}=\left [ p+6\sqrt{q}+4q^{\frac{1}{4}}\sqrt{p+2\sqrt{q}} \right ]^{\frac{1}{4}}\)
