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If-the-ratio-of-the-roots-of-ax-2-bx-b-0-is-p-q-then-show-that-p-q-q-p-b-a-0-

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Question Number 202392 by MATHEMATICSAM last updated on 26/Dec/23
If the ratio of the roots of ax^2 + bx + b = 0 is p : q then show that (√(p/q)) + (√(q/p)) + (√(b/a)) = 0.
$$\mathrm{If}\:\mathrm{the}\:\mathrm{ratio}\:\mathrm{of}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:{ax}^{\mathrm{2}} \:+\:{bx}\:+\:{b}\:=\:\mathrm{0} \\ $$$$\mathrm{is}\:{p}\::\:{q}\:\mathrm{then}\:\mathrm{show}\:\mathrm{that} \\ $$$$\sqrt{\frac{{p}}{{q}}}\:+\:\sqrt{\frac{{q}}{{p}}}\:+\:\sqrt{\frac{{b}}{{a}}}\:=\:\mathrm{0}. \\ $$
Commented by MATHEMATICSAM last updated on 26/Dec/23
The question is corrected now
$$\mathrm{The}\:\mathrm{question}\:\mathrm{is}\:\mathrm{corrected}\:\mathrm{now} \\ $$
Commented by Frix last updated on 27/Dec/23
Example 2x^2 +9x+9=0 p=−3∧q=−(3/2) (√(p/q))+(√(q/p))+(√(b/a))=3(√2)
$$\mathrm{Example} \\ $$$$\mathrm{2}{x}^{\mathrm{2}} +\mathrm{9}{x}+\mathrm{9}=\mathrm{0} \\ $$$${p}=−\mathrm{3}\wedge{q}=−\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\sqrt{\frac{{p}}{{q}}}+\sqrt{\frac{{q}}{{p}}}+\sqrt{\frac{{b}}{{a}}}=\mathrm{3}\sqrt{\mathrm{2}} \\ $$
Commented by MATHEMATICSAM last updated on 27/Dec/23
Let α and β are the roots of ax^2 + bx + b = 0 So α + β = − (b/a) , αβ = (b/a) , (p/q) = (α/β) (√(p/q)) + (√(q/p)) + (√(b/a)) = (√(α/β)) + (√(β/α)) + (√(b/a)) = ((α + β)/( (√(αβ)))) + (√(b/a)) = ((− (b/a))/( (√(b/a)))) + (√(b/a)) = − (√(b/a)) + (√(b/a)) = 0 (Proved)
$$\mathrm{Let}\:\alpha\:\mathrm{and}\:\beta\:\mathrm{are}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:{ax}^{\mathrm{2}} \:+\:{bx}\:+\:{b}\:=\:\mathrm{0} \\ $$$$\mathrm{So}\:\alpha\:+\:\beta\:=\:−\:\frac{{b}}{{a}}\:,\:\alpha\beta\:=\:\frac{{b}}{{a}}\:,\:\frac{{p}}{{q}}\:=\:\frac{\alpha}{\beta} \\ $$$$\sqrt{\frac{{p}}{{q}}}\:+\:\sqrt{\frac{{q}}{{p}}}\:+\:\sqrt{\frac{{b}}{{a}}}\: \\ $$$$=\:\sqrt{\frac{\alpha}{\beta}}\:+\:\sqrt{\frac{\beta}{\alpha}}\:+\:\sqrt{\frac{{b}}{{a}}} \\ $$$$=\:\frac{\alpha\:+\:\beta}{\:\sqrt{\alpha\beta}}\:+\:\sqrt{\frac{{b}}{{a}}} \\ $$$$=\:\frac{−\:\frac{{b}}{{a}}}{\:\sqrt{\frac{{b}}{{a}}}}\:+\:\sqrt{\frac{{b}}{{a}}} \\ $$$$=\:−\:\sqrt{\frac{{b}}{{a}}}\:+\:\sqrt{\frac{{b}}{{a}}} \\ $$$$=\:\mathrm{0}\:\left(\mathrm{Proved}\right) \\ $$
Commented by MATHEMATICSAM last updated on 27/Dec/23
not (c/a) Equation is given ax^2 + bx + b = 0
$$\mathrm{not}\:\frac{{c}}{{a}}\:\mathrm{Equation}\:\mathrm{is}\:\mathrm{given}\:{ax}^{\mathrm{2}} \:+\:{bx}\:+\:{b}\:=\:\mathrm{0} \\ $$
Commented by Rasheed.Sindhi last updated on 27/Dec/23
Pl check carefully: = ((α + β)/( (√(αβ)))) + (√(b/a)) = ((−(b/a))/( (√(c/a)))) + (√(b/a))
$${Pl}\:{check}\:{carefully}: \\ $$$$=\:\frac{\alpha\:+\:\beta}{\:\sqrt{\alpha\beta}}\:+\:\sqrt{\frac{{b}}{{a}}}\: \\ $$$$=\:\frac{−\frac{{b}}{{a}}}{\:\sqrt{\frac{{c}}{{a}}}}\:+\:\sqrt{\frac{{b}}{{a}}} \\ $$$$ \\ $$
Commented by Rasheed.Sindhi last updated on 27/Dec/23
Ok
$${Ok} \\ $$
Commented by Frix last updated on 27/Dec/23
(√(α/β))+(√(β/α))≠((α+β)/( (√(αβ)))) α=−4∧β=−9 (√(α/β))+(√(β/α))=((13)/6) but ((α+β)/( (√(αβ))))=−((13)/6) α=−4∧β=9 (√(α/β))+(√(β/α))=((13)/6)i but ((α+β)/( (√(αβ))))=−(5/6)i
$$\sqrt{\frac{\alpha}{\beta}}+\sqrt{\frac{\beta}{\alpha}}\neq\frac{\alpha+\beta}{\:\sqrt{\alpha\beta}} \\ $$$$\alpha=−\mathrm{4}\wedge\beta=−\mathrm{9} \\ $$$$\sqrt{\frac{\alpha}{\beta}}+\sqrt{\frac{\beta}{\alpha}}=\frac{\mathrm{13}}{\mathrm{6}}\:\mathrm{but}\:\frac{\alpha+\beta}{\:\sqrt{\alpha\beta}}=−\frac{\mathrm{13}}{\mathrm{6}} \\ $$$$\alpha=−\mathrm{4}\wedge\beta=\mathrm{9} \\ $$$$\sqrt{\frac{\alpha}{\beta}}+\sqrt{\frac{\beta}{\alpha}}=\frac{\mathrm{13}}{\mathrm{6}}\mathrm{i}\:\mathrm{but}\:\frac{\alpha+\beta}{\:\sqrt{\alpha\beta}}=−\frac{\mathrm{5}}{\mathrm{6}}\mathrm{i} \\ $$
Commented by Frix last updated on 27/Dec/23
You can ignore this but then it′s not mathematics.
$$\mathrm{You}\:\mathrm{can}\:\mathrm{ignore}\:\mathrm{this}\:\mathrm{but}\:\mathrm{then}\:\mathrm{it}'\mathrm{s}\:\mathrm{not} \\ $$$$\mathrm{mathematics}. \\ $$
Commented by Rasheed.Sindhi last updated on 27/Dec/23
Sir, you′re very right!
$$\boldsymbol{\mathrm{Sir}},\:\mathrm{you}'\mathrm{re}\:\mathrm{very}\:\mathrm{right}! \\ $$
Commented by MATHEMATICSAM last updated on 27/Dec/23
Hmm right but for this question made by basic algebric calculations it is correct. Basically for negative numbers the square root rules sometimes do not work.
$$\mathrm{Hmm}\:\mathrm{right}\:\mathrm{but}\:\mathrm{for}\:\mathrm{this}\:\mathrm{question}\:\mathrm{made} \\ $$$$\mathrm{by}\:\mathrm{basic}\:\mathrm{algebric}\:\mathrm{calculations}\:\mathrm{it}\:\mathrm{is}\:\mathrm{correct}. \\ $$$$\mathrm{Basically}\:\mathrm{for}\:\mathrm{negative}\:\mathrm{numbers}\:\mathrm{the}\:\mathrm{square} \\ $$$$\mathrm{root}\:\mathrm{rules}\:\mathrm{sometimes}\:\mathrm{do}\:\mathrm{not}\:\mathrm{work}. \\ $$
Commented by Frix last updated on 27/Dec/23
If it′s wrong it′s wrong. You′d have to restrict the values to get a valid solution.
$$\mathrm{If}\:\mathrm{it}'\mathrm{s}\:\mathrm{wrong}\:\mathrm{it}'\mathrm{s}\:\mathrm{wrong}. \\ $$$$\mathrm{You}'\mathrm{d}\:\mathrm{have}\:\mathrm{to}\:\mathrm{restrict}\:\mathrm{the}\:\mathrm{values}\:\mathrm{to}\:\mathrm{get} \\ $$$$\mathrm{a}\:\mathrm{valid}\:\mathrm{solution}. \\ $$
Answered by Rasheed.Sindhi last updated on 27/Dec/23
Let α & β are roots α+β=−b/a , αβ=c/a (√(α/β)) +(√(β/α)) +(√(b/a)) =0 ((√α)/( (√β)))+((√β)/( (√α)))+(√(b/a)) =0 ((α+β)/( (√(αβ))))+(√(b/a)) =0 ((−b/a)/( (√(b/a))))+(√(b/a)) =0 ((((−b)/a)+(b/a))/( (√(b/a))))=0 (0/( (√(b/a))))=0 0=0 Proved
$${Let}\:\alpha\:\&\:\beta\:{are}\:{roots} \\ $$$$\alpha+\beta=−{b}/{a}\:,\:\alpha\beta={c}/{a} \\ $$$$\sqrt{\frac{\alpha}{\beta}}\:+\sqrt{\frac{\beta}{\alpha}}\:+\sqrt{\frac{{b}}{{a}}}\:=\mathrm{0} \\ $$$$\frac{\sqrt{\alpha}}{\:\sqrt{\beta}}+\frac{\sqrt{\beta}}{\:\sqrt{\alpha}}+\sqrt{\frac{{b}}{{a}}}\:=\mathrm{0} \\ $$$$\frac{\alpha+\beta}{\:\sqrt{\alpha\beta}}+\sqrt{\frac{{b}}{{a}}}\:=\mathrm{0} \\ $$$$\frac{−{b}/{a}}{\:\sqrt{\frac{{b}}{{a}}}}+\sqrt{\frac{{b}}{{a}}}\:=\mathrm{0} \\ $$$$\frac{\cancel{\frac{−{b}}{{a}}}+\cancel{\frac{{b}}{{a}}}}{\:\sqrt{\frac{{b}}{{a}}}}=\mathrm{0} \\ $$$$\frac{\mathrm{0}}{\:\sqrt{\frac{{b}}{{a}}}}=\mathrm{0} \\ $$$$\mathrm{0}=\mathrm{0}\:\mathbb{P}\boldsymbol{\mathrm{roved}} \\ $$
Answered by Frix last updated on 26/Dec/23
(p/q)∈R^+ ⇒ (√(p/q))>0∧(√(q/p))>0 (c/a)∈R^+ ⇒ (√(c/a))>0 ⇒ (√(p/q))+(√(q/p))+(√(c/a))≠0 (p/q)=0 ⇒ (q/p) not defined (p/q)∈R^− ⇒ −(p/q)∈R^+ ⇒ ⇒(√(p/q))=i(√(−(p/q)))∧(√(q/p))=i(√(−(q/p))) ⇒ ⇒ (√(p/q))+(√(q/p))=iu∧u>0 (√(c/a))=v∨iv∧v∈R^+ ⇒ (√(p/q))+(√(q/p))+(√(c/a))≠0
$$\frac{{p}}{{q}}\in\mathbb{R}^{+} \:\Rightarrow\:\sqrt{\frac{{p}}{{q}}}>\mathrm{0}\wedge\sqrt{\frac{{q}}{{p}}}>\mathrm{0} \\ $$$$\frac{{c}}{{a}}\in\mathbb{R}^{+} \:\Rightarrow\:\sqrt{\frac{{c}}{{a}}}>\mathrm{0} \\ $$$$\Rightarrow\:\sqrt{\frac{{p}}{{q}}}+\sqrt{\frac{{q}}{{p}}}+\sqrt{\frac{{c}}{{a}}}\neq\mathrm{0} \\ $$$$ \\ $$$$\frac{{p}}{{q}}=\mathrm{0}\:\Rightarrow\:\frac{{q}}{{p}}\:\mathrm{not}\:\mathrm{defined} \\ $$$$ \\ $$$$\frac{{p}}{{q}}\in\mathbb{R}^{−} \:\Rightarrow\:−\frac{{p}}{{q}}\in\mathbb{R}^{+} \:\Rightarrow\: \\ $$$$\Rightarrow\sqrt{\frac{{p}}{{q}}}=\mathrm{i}\sqrt{−\frac{{p}}{{q}}}\wedge\sqrt{\frac{{q}}{{p}}}=\mathrm{i}\sqrt{−\frac{{q}}{{p}}}\:\Rightarrow \\ $$$$\Rightarrow\:\sqrt{\frac{{p}}{{q}}}+\sqrt{\frac{{q}}{{p}}}=\mathrm{i}{u}\wedge{u}>\mathrm{0} \\ $$$$\sqrt{\frac{{c}}{{a}}}={v}\vee\mathrm{i}{v}\wedge{v}\in\mathbb{R}^{+} \\ $$$$\Rightarrow\:\sqrt{\frac{{p}}{{q}}}+\sqrt{\frac{{q}}{{p}}}+\sqrt{\frac{{c}}{{a}}}\neq\mathrm{0} \\ $$

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