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Question Number 202392 by MATHEMATICSAM last updated on 26/Dec/23

$$\mathrm{If}\mathrm{the}\mathrm{ratio}\mathrm{of}\mathrm{the}\mathrm{roots}\mathrm{of}{ax}^2+bx+b=0$$$$\mathrm{is}p:q\mathrm{then}\mathrm{show}\mathrm{that}$$$$\sqrt{\frac{p}{q}}+\sqrt{\frac{q}{p}}+\sqrt{\frac{b}{a}}=0.$$

Commented by MATHEMATICSAM last updated on 26/Dec/23

$$\mathrm{The}\mathrm{question}\mathrm{is}\mathrm{corrected}\mathrm{now}$$

Commented by Frix last updated on 27/Dec/23

$$\mathrm{Example}$$$$2x^2+9x+9=0$$$$p=-3\wedge q=-\frac{3}{2}$$$$\sqrt{\frac{p}{q}}+\sqrt{\frac{q}{p}}+\sqrt{\frac{b}{a}}=3\sqrt{2}$$

Commented by MATHEMATICSAM last updated on 27/Dec/23

$$\mathrm{Let}\alpha \mathrm{and}\beta \mathrm{are}\mathrm{the}\mathrm{roots}\mathrm{of}{ax}^2+bx+b=0$$$$\mathrm{So}\alpha +\beta =-\frac{b}{a},\alpha \beta =\frac{b}{a},\frac{p}{q}=\frac{\alpha }{\beta }$$$$\sqrt{\frac{p}{q}}+\sqrt{\frac{q}{p}}+\sqrt{\frac{b}{a}}$$$$=\sqrt{\frac{\alpha }{\beta }}+\sqrt{\frac{\beta }{\alpha }}+\sqrt{\frac{b}{a}}$$$$=\frac{\alpha +\beta }{\sqrt{\alpha \beta }}+\sqrt{\frac{b}{a}}$$$$=\frac{-\frac{b}{a}}{\sqrt{\frac{b}{a}}}+\sqrt{\frac{b}{a}}$$$$=-\sqrt{\frac{b}{a}}+\sqrt{\frac{b}{a}}$$$$=0\left(\mathrm{Proved}\right)$$

Commented by MATHEMATICSAM last updated on 27/Dec/23

$$\mathrm{not}\frac{c}{a}\mathrm{Equation}\mathrm{is}\mathrm{given}{ax}^2+bx+b=0$$

Commented by Rasheed.Sindhi last updated on 27/Dec/23

$$Plcheckcarefully:$$$$=\frac{\alpha +\beta }{\sqrt{\alpha \beta }}+\sqrt{\frac{b}{a}}$$$$=\frac{-\frac{b}{a}}{\sqrt{\frac{c}{a}}}+\sqrt{\frac{b}{a}}$$$$$$

Commented by Rasheed.Sindhi last updated on 27/Dec/23

$$Ok$$

Commented by Frix last updated on 27/Dec/23

$$\sqrt{\frac{\alpha }{\beta }}+\sqrt{\frac{\beta }{\alpha }}\ne \frac{\alpha +\beta }{\sqrt{\alpha \beta }}$$$$\alpha =-4\wedge \beta =-9$$$$\sqrt{\frac{\alpha }{\beta }}+\sqrt{\frac{\beta }{\alpha }}=\frac{13}{6}\mathrm{but}\frac{\alpha +\beta }{\sqrt{\alpha \beta }}=-\frac{13}{6}$$$$\alpha =-4\wedge \beta =9$$$$\sqrt{\frac{\alpha }{\beta }}+\sqrt{\frac{\beta }{\alpha }}=\frac{13}{6}\mathrm{i}\mathrm{but}\frac{\alpha +\beta }{\sqrt{\alpha \beta }}=-\frac{5}{6}\mathrm{i}$$

Commented by Frix last updated on 27/Dec/23

$$\mathrm{You}\mathrm{can}\mathrm{ignore}\mathrm{this}\mathrm{but}\mathrm{then}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{not}$$$$\mathrm{mathematics}.$$

Commented by Rasheed.Sindhi last updated on 27/Dec/23

$$\mathbf{Sir},{\mathrm{you}}^{\prime }\mathrm{re}\mathrm{very}\mathrm{right}!$$

Commented by MATHEMATICSAM last updated on 27/Dec/23

$$\mathrm{Hmm}\mathrm{right}\mathrm{but}\mathrm{for}\mathrm{this}\mathrm{question}\mathrm{made}$$$$\mathrm{by}\mathrm{basic}\mathrm{algebric}\mathrm{calculations}\mathrm{it}\mathrm{is}\mathrm{correct}.$$$$\mathrm{Basically}\mathrm{for}\mathrm{negative}\mathrm{numbers}\mathrm{the}\mathrm{square}$$$$\mathrm{root}\mathrm{rules}\mathrm{sometimes}\mathrm{do}\mathrm{not}\mathrm{work}.$$

Commented by Frix last updated on 27/Dec/23

$$\mathrm{If}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{wrong}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{wrong}.$$$${\mathrm{You}}^{\prime }\mathrm{d}\mathrm{have}\mathrm{to}\mathrm{restrict}\mathrm{the}\mathrm{values}\mathrm{to}\mathrm{get}$$$$\mathrm{a}\mathrm{valid}\mathrm{solution}.$$

Answered by Rasheed.Sindhi last updated on 27/Dec/23

$$Let\alpha \mathrm{\&}\beta areroots$$$$\alpha +\beta =-b/a,\alpha \beta =c/a$$$$\sqrt{\frac{\alpha }{\beta }}+\sqrt{\frac{\beta }{\alpha }}+\sqrt{\frac{b}{a}}=0$$$$\frac{\sqrt{\alpha }}{\sqrt{\beta }}+\frac{\sqrt{\beta }}{\sqrt{\alpha }}+\sqrt{\frac{b}{a}}=0$$$$\frac{\alpha +\beta }{\sqrt{\alpha \beta }}+\sqrt{\frac{b}{a}}=0$$$$\frac{-b/a}{\sqrt{\frac{b}{a}}}+\sqrt{\frac{b}{a}}=0$$$$\frac{\cancel{\frac{-b}{a}}+\cancel{\frac{b}{a}}}{\sqrt{\frac{b}{a}}}=0$$$$\frac{0}{\sqrt{\frac{b}{a}}}=0$$$$0=0\mathbb{P}\mathbf{roved}$$

Answered by Frix last updated on 26/Dec/23

$$\frac{p}{q}\in {\mathbb{R}}^{+}\Rightarrow \sqrt{\frac{p}{q}}>0\wedge \sqrt{\frac{q}{p}}>0$$$$\frac{c}{a}\in {\mathbb{R}}^{+}\Rightarrow \sqrt{\frac{c}{a}}>0$$$$\Rightarrow \sqrt{\frac{p}{q}}+\sqrt{\frac{q}{p}}+\sqrt{\frac{c}{a}}\ne 0$$$$$$$$\frac{p}{q}=0\Rightarrow \frac{q}{p}\mathrm{not}\mathrm{defined}$$$$$$$$\frac{p}{q}\in {\mathbb{R}}^{-}\Rightarrow -\frac{p}{q}\in {\mathbb{R}}^{+}\Rightarrow$$$$\Rightarrow \sqrt{\frac{p}{q}}=\mathrm{i}\sqrt{-\frac{p}{q}}\wedge \sqrt{\frac{q}{p}}=\mathrm{i}\sqrt{-\frac{q}{p}}\Rightarrow$$$$\Rightarrow \sqrt{\frac{p}{q}}+\sqrt{\frac{q}{p}}=\mathrm{i}u\wedge u>0$$$$\sqrt{\frac{c}{a}}=v\vee \mathrm{i}v\wedge v\in {\mathbb{R}}^{+}$$$$\Rightarrow \sqrt{\frac{p}{q}}+\sqrt{\frac{q}{p}}+\sqrt{\frac{c}{a}}\ne 0$$

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