Question-59226

Question Number 59226 by azizullah last updated on 06/May/19

Answered by MJS last updated on 06/May/19

trying with x∈Z at first (√(−x)) ∈Z ⇒ x=−z^2 x=−1 ⇒ wrong x=−4 ⇒ wrong x=−9 ⇒ true 2×(−9)+5=−13

$$\mathrm{trying}\:\mathrm{with}\:{x}\in\mathbb{Z}\:\mathrm{at}\:\mathrm{first} \\ $$$$\sqrt{−{x}}\:\in\mathbb{Z}\:\Rightarrow\:{x}=−{z}^{\mathrm{2}} \\ $$$${x}=−\mathrm{1}\:\Rightarrow\:\mathrm{wrong} \\ $$$${x}=−\mathrm{4}\:\Rightarrow\:\mathrm{wrong} \\ $$$${x}=−\mathrm{9}\:\Rightarrow\:\mathrm{true} \\ $$$$\mathrm{2}×\left(−\mathrm{9}\right)+\mathrm{5}=−\mathrm{13} \\ $$

Commented by azizullah last updated on 07/May/19

dear sir please solve by long radical equation method.

$$\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{dear}}\:\boldsymbol{\mathrm{sir}}\:\boldsymbol{\mathrm{please}}\:\boldsymbol{\mathrm{solve}}\:\boldsymbol{\mathrm{by}}\:\boldsymbol{\mathrm{long}}\:\boldsymbol{\mathrm{radical}}\:\boldsymbol{\mathrm{equation}}\:\boldsymbol{\mathrm{method}}. \\ $$

Answered by MJS last updated on 07/May/19

(√(40−9x))=(√(−x))+2(√(7−x)) ∣^2 40−9x=28−5x+4(√(−x))(√(7−x)) 3−x=(√(−x))(√(7−x)) ∣^2 x^2 −6x+9=x^2 −7x x+9=0 x=−9

$$\sqrt{\mathrm{40}−\mathrm{9}{x}}=\sqrt{−{x}}+\mathrm{2}\sqrt{\mathrm{7}−{x}}\:\:\:\:\:\mid^{\mathrm{2}} \\ $$$$\mathrm{40}−\mathrm{9}{x}=\mathrm{28}−\mathrm{5}{x}+\mathrm{4}\sqrt{−{x}}\sqrt{\mathrm{7}−{x}} \\ $$$$\mathrm{3}−{x}=\sqrt{−{x}}\sqrt{\mathrm{7}−{x}}\:\:\:\:\:\mid^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{9}={x}^{\mathrm{2}} −\mathrm{7}{x} \\ $$$${x}+\mathrm{9}=\mathrm{0} \\ $$$${x}=−\mathrm{9} \\ $$

Commented by azizullah last updated on 07/May/19

$$\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{dear}}\:\boldsymbol{\mathrm{sir}}\:\boldsymbol{\mathrm{required}}\:\boldsymbol{\mathrm{answer}}\:\boldsymbol{\mathrm{is}}\:\mathrm{3} \\ $$

Commented by MJS last updated on 07/May/19

(√(40−9x))−2(√(7−x))=(√(−x)) x=3 ⇒ (√(13))−4=(√3)i which is wrong 2x+5=3 ⇒ x=−1 ⇒ 7−4(√2)=1 which is wrong x=−9 ⇒ 11−8=3 which is true 2x+5=−13

$$\sqrt{\mathrm{40}−\mathrm{9}{x}}−\mathrm{2}\sqrt{\mathrm{7}−{x}}=\sqrt{−{x}} \\ $$$${x}=\mathrm{3}\:\Rightarrow\:\sqrt{\mathrm{13}}−\mathrm{4}=\sqrt{\mathrm{3}}\mathrm{i}\:\mathrm{which}\:\mathrm{is}\:\mathrm{wrong} \\ $$$$\mathrm{2}{x}+\mathrm{5}=\mathrm{3}\:\Rightarrow\:{x}=−\mathrm{1}\:\Rightarrow\:\mathrm{7}−\mathrm{4}\sqrt{\mathrm{2}}=\mathrm{1}\:\mathrm{which}\:\mathrm{is}\:\mathrm{wrong} \\ $$$${x}=−\mathrm{9}\:\Rightarrow\:\mathrm{11}−\mathrm{8}=\mathrm{3}\:\mathrm{which}\:\mathrm{is}\:\mathrm{true} \\ $$$$\mathrm{2}{x}+\mathrm{5}=−\mathrm{13} \\ $$

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