An algebra problem by حسن العطية

Algebra Level 2

If a 2 − 1 b 2 b 2 − 1 a 2 = 9 \dfrac { { a }^{ 2 }-\dfrac { 1 }{ { b }^{ 2 } } }{ { b }^{ 2 }-\dfrac { 1 }{ { a }^{ 2 } } } =9 , then what is the value of ∣ a b ∣ \left| \dfrac { a }{ b } \right| ?

and ∣ a b ∣ ≠ 1 \left| ab \right| \neq 1


The answer is 3.

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حسن العطية by حسن العطية , 22, Saudi Arabia

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2 solutions

a 2 − 1 b 2 b 2 − 1 a 2 = 9 \frac { { a }^{ 2 }-\frac { 1 }{ { b }^{ 2 } } }{ { b }^{ 2 }-\frac { 1 }{ { a }^{ 2 } } } =9

b 2 a 2 − 1 b 2 a 2 b 2 − 1 a 2 = 9 \frac { \frac { { b }^{ 2 }{ a }^{ 2 }-1 }{ { b }^{ 2 } } }{ \frac { { a }^{ 2 }{ b }^{ 2 }-1 }{ { a }^{ 2 } } } =9

a 2 ( b 2 a 2 − 1 ) b 2 ( a 2 b 2 − 1 ) = 9 \frac { { a }^{ 2 }({ b }^{ 2 }{ a }^{ 2 }-1) }{ { b }^{ 2 }({ a }^{ 2 }{ b }^{ 2 }-1) } =9

a 2 b 2 = 9 \frac { { a }^{ 2 } }{ { b }^{ 2 } } =9

a 2 b 2 = 9 \sqrt { \frac { { a }^{ 2 } }{ { b }^{ 2 } } } =\sqrt { 9 }

a b = ± 3 \frac { { a } }{ { b } } =\pm 3

∣ a b ∣ = ∣ ± 3 ∣ \left| \frac { { a } }{ { b } } \right| =\left| \pm 3 \right|

∣ a b ∣ = 3 \left| \frac { { a } }{ { b } } \right| =3

3 \boxed{3}

4 Helpful 0 Interesting 0 Brilliant 0 Confused

Please mention in the question that ∣ a b ∣ ≠ 1 |ab|\ne1 . That would make it more clear.

Vignesh S - 5 years, 1 month ago

a 2 − 1 b 2 b 2 − 1 a 2 = a 2 ( 1 − 1 a 2 b 2 ) b 2 ( 1 − 1 a 2 b 2 ) = a 2 b 2 = 9 \dfrac{a^2-\dfrac 1 {b^2}}{b^2-\dfrac 1 {a^2}}=\dfrac{a^2 \left(1-\dfrac 1 {a^2b^2} \right)}{b^2 \left(1-\dfrac 1 {a^2b^2} \right)}=\dfrac{a^2}{b^2}=9 .

So, ∣ a b ∣ = 9 = 3 \left|\dfrac{a}{b}\right|=\sqrt{9}=3 .

0 Helpful 0 Interesting 0 Brilliant 0 Confused

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