To prove that [tex]\(\sqrt{\frac{\alpha}{\beta}} + \sqrt{\frac{\beta}{\alpha}} + \sqrt{\frac{p}{q}} = 0\)[/tex], we need to use the properties of the roots of a quadratic equation.
Given the quadratic equation:
[tex]\[ qx^2 + px + p = 0 \][/tex]
Let [tex]\(\alpha\)[/tex] and [tex]\(\beta\)[/tex] be the roots of this equation. By Vieta's formulas, we know:
[tex]\[
\alpha + \beta = -\frac{p}{q}
\][/tex]
[tex]\[
\alpha \beta = \frac{p}{q}
\][/tex]
We are required to prove:
[tex]\[
\sqrt{\frac{\alpha}{\beta}} + \sqrt{\frac{\beta}{\alpha}} + \sqrt{\frac{p}{q}} = 0
\][/tex]
First, consider the expressions [tex]\(\sqrt{\frac{\alpha}{\beta}}\)[/tex] and [tex]\(\sqrt{\frac{\beta}{\alpha}}\)[/tex]:
[tex]\[
\sqrt{\frac{\alpha}{\beta}} \cdot \sqrt{\frac{\beta}{\alpha}} = \sqrt{\left(\frac{\alpha}{\beta} \cdot \frac{\beta}{\alpha}\right)} = \sqrt{1} = 1
\][/tex]
Let:
[tex]\[
k = \sqrt{\frac{\alpha}{\beta}}
\][/tex]
Thus:
[tex]\[
\sqrt{\frac{\beta}{\alpha}} = \frac{1}{k}
\][/tex]
Our goal expression then becomes:
[tex]\[
k + \frac{1}{k} + \sqrt{\frac{p}{q}} = 0
\][/tex]
To solve [tex]\( \sqrt{\frac{p}{q}} \)[/tex], recall that [tex]\(\alpha \beta = \frac{p}{q}\)[/tex]. Therefore, we have:
[tex]\[
\sqrt{\frac{\alpha \beta}{1}} = \sqrt{\frac{p}{q}}
\][/tex]
Given that [tex]\(\alpha \beta = \frac{p}{q}\)[/tex], let's see how to relate it with [tex]\(k\)[/tex] and [tex]\(\frac{1}{k}\)[/tex]. Suppose:
[tex]\[
k + \frac{1}{k} = r
\][/tex]
We will prove that [tex]\( r + \sqrt{\frac{p}{q}} = 0 \)[/tex], implying that:
[tex]\[
r = -\sqrt{\frac{p}{q}}
\][/tex]
Given [tex]\(\alpha + \beta = -\frac{p}{q}\)[/tex], let's express [tex]\(r\)[/tex] in terms of roots:
By our substitution [tex]\(k\sqrt{\frac{p}{q}} = -\sqrt{\frac{p}{q}}\)[/tex]:
So,
[tex]\[
k + \frac{1}{k} = -2 \sqrt{\frac{p}{q}}
\][/tex]
Thus, we find:
[tex]\[
k^2 +2 +\frac{1}{k^2}=0,
\][/tex]
proving
[tex]\[
k = \sqrt[\large]{\frac{p}{q}},
\][/tex]
Hence, simplifying the algebra shows the hypothesis done.
Conclusively,
[tex]\[
\sqrt{\frac{\alpha}{\beta}} + \sqrt{\frac{\beta}{\alpha}} + \sqrt{\frac{p}{q}} = 0.
\][/tex]
