If p and q are the roots of the equation $lx^2+nx+n=0$, show that $\sqrt{\dfrac{p}{q}}+\sqrt{\dfrac{q}{p}}+\sqrt{\dfrac{n}{l}}=0$

> [!TLDR] > Expresses $\sqrt{p/q}+\sqrt{q/p}$ as $(p+q)/\sqrt{pq}$ and substitutes the sum and product of roots from $lx^2+nx+n=0$. > > 1. Since $p,q$ are roots of $lx^2+nx+n=0$: sum of roots $p+q=-\dfrac{n}{l}$, product of roots $pq=\dfrac{n}{l}$. > 2. Combine the surds: $\sqrt{\dfrac pq}+\sqrt{\dfrac qp}=\dfrac{\sqrt p}{\sqrt q}+\dfrac{\sqrt q}{\sqrt p}=\dfrac{(\sqrt p)^2+(\sqrt q)^2}{\sqrt p\sqrt q}=\dfrac{p+q}{\sqrt{pq}}$. > 3. Substitute: $\dfrac{p+q}{\sqrt{pq}}=\dfrac{-n/l}{\sqrt{n/l}}=-\dfrac{n/l}{\sqrt{n/l}}=-\sqrt{\dfrac nl}$ (since $\dfrac{A}{\sqrt A}=\sqrt A$). …

If p and q are the roots of the equation lx²+nx+n=0, show that √((p)/(q))+√((q)/(p))+√((n)/(l))=0

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lx^2+nx+n=0

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Expresses $\sqrt{p/q}+\sqrt{q/p}$ as $(p+q)/\sqrt{pq}$ and substitutes the sum and product of roots from $lx^2+nx+n=0$ .

Since $p,q$ are roots of $lx^2+nx+n=0$ : sum of roots $p+q=-\dfrac{n}{l}$ , product of roots $pq=\dfrac{n}{l}$ .
Combine the surds: $\sqrt{\dfrac pq}+\sqrt{\dfrac qp}=\dfrac{\sqrt p}{\sqrt q}+\dfrac{\sqrt q}{\sqrt p}=\dfrac{(\sqrt p)^2+(\sqrt q)^2}{\sqrt p\sqrt q}=\dfrac{p+q}{\sqrt{pq}}$ .
Substitute: $\dfrac{p+q}{\sqrt{pq}}=\dfrac{-n/l}{\sqrt{n/l}}=-\dfrac{n/l}{\sqrt{n/l}}=-\sqrt{\dfrac nl}$ (since $\dfrac{A}{\sqrt A}=\sqrt A$ ). …

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Q.If p and q are the roots of the equation $lx^2+nx+n=0$ , show that $\sqrt{\dfrac{p}{q}}+\sqrt{\dfrac{q}{p}}+\sqrt{\dfrac{n}{l}}=0$

Concept understanding — Relations between Roots and Coefficients (Vieta's Formulas)

Q.If p and q are the roots of the equation lx2+nx+n=0lx^2+nx+n=0lx2+nx+n=0, show that pq+qp+nl=0\sqrt{\dfrac{p}{q}}+\sqrt{\dfrac{q}{p}}+\sqrt{\dfrac{n}{l}}=0qp​​+pq​​+ln​​=0

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Q.If p and q are the roots of the equation $lx^2+nx+n=0$ , show that $\sqrt{\dfrac{p}{q}}+\sqrt{\dfrac{q}{p}}+\sqrt{\dfrac{n}{l}}=0$

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