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Question Number 112805 by Aina Samuel Temidayo last updated on 09/Sep/20

If x=(((√(a+2b))+ (√(a−2b)))/( (√(a+2b))− (√(a−2b)))), then the value of bx^2 −ax+b is

$$\mathrm{If}\:\mathrm{x}=\frac{\sqrt{\mathrm{a}+\mathrm{2b}}+\:\sqrt{\mathrm{a}−\mathrm{2b}}}{\:\sqrt{\mathrm{a}+\mathrm{2b}}−\:\sqrt{\mathrm{a}−\mathrm{2b}}},\:\mathrm{then}\:\mathrm{the}\:\mathrm{value} \\ $$$$\mathrm{of}\:\mathrm{bx}^{\mathrm{2}} −\mathrm{ax}+\mathrm{b}\:\mathrm{is} \\ $$

Answered by nimnim last updated on 09/Sep/20

(x/1)=(((√(a+2b))+(√(a−2b)))/( (√(a+2b))−(√(a−2b)))) ⇒((x+1)/(x−1))=((2(√(a+2b)))/( 2(√(a−2b)))) (by componendo & dividendo) ⇒((x^2 +2x+1)/(x^2 −2x+1))=((a+2b)/(a−2b)) ⇒((2(x^2 +1))/(4x))=((2a)/(4b)) (by componendo & dividendo) ⇒((x^2 +1)/x)=(a/b) ⇒bx^2 +b=ax ⇒bx^2 −ax+b=0★

$$\:\:\:\:\:\frac{\mathrm{x}}{\mathrm{1}}=\frac{\sqrt{\mathrm{a}+\mathrm{2b}}+\sqrt{\mathrm{a}−\mathrm{2b}}}{\:\sqrt{\mathrm{a}+\mathrm{2b}}−\sqrt{\mathrm{a}−\mathrm{2b}}} \\ $$$$\Rightarrow\frac{\mathrm{x}+\mathrm{1}}{\mathrm{x}−\mathrm{1}}=\frac{\mathrm{2}\sqrt{\mathrm{a}+\mathrm{2b}}}{\:\mathrm{2}\sqrt{\mathrm{a}−\mathrm{2b}}}\:\left(\mathrm{by}\:\mathrm{componendo}\:\&\:\mathrm{dividendo}\right) \\ $$$$\Rightarrow\frac{\mathrm{x}^{\mathrm{2}} +\mathrm{2x}+\mathrm{1}}{\mathrm{x}^{\mathrm{2}} −\mathrm{2x}+\mathrm{1}}=\frac{\mathrm{a}+\mathrm{2b}}{\mathrm{a}−\mathrm{2b}} \\ $$$$\Rightarrow\frac{\mathrm{2}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)}{\mathrm{4x}}=\frac{\mathrm{2a}}{\mathrm{4b}}\:\:\left(\mathrm{by}\:\mathrm{componendo}\:\&\:\mathrm{dividendo}\right) \\ $$$$\Rightarrow\frac{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}{\mathrm{x}}=\frac{\mathrm{a}}{\mathrm{b}} \\ $$$$\Rightarrow\mathrm{bx}^{\mathrm{2}} +\mathrm{b}=\mathrm{ax} \\ $$$$\Rightarrow\mathrm{bx}^{\mathrm{2}} −\mathrm{ax}+\mathrm{b}=\mathrm{0}\bigstar \\ $$

Commented by Aina Samuel Temidayo last updated on 09/Sep/20

Ok. Thanks.

$$\mathrm{Ok}.\:\mathrm{Thanks}. \\ $$

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