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Question Number 71242 by TawaTawa last updated on 13/Oct/19

Given: (a/b) + (c/d) = (b/a) + (d/c) Show that, (a^2 /b^2 ) − (c^2 /d^2 ) = (b^2 /a^2 ) − (d^2 /c^2 )

$$\mathrm{Given}:\:\:\:\frac{\mathrm{a}}{\mathrm{b}}\:+\:\frac{\mathrm{c}}{\mathrm{d}}\:\:=\:\:\frac{\mathrm{b}}{\mathrm{a}}\:+\:\frac{\mathrm{d}}{\mathrm{c}} \\ $$$$\mathrm{Show}\:\mathrm{that},\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{a}^{\mathrm{2}} }{\mathrm{b}^{\mathrm{2}} }\:−\:\frac{\mathrm{c}^{\mathrm{2}} }{\mathrm{d}^{\mathrm{2}} }\:\:=\:\:\frac{\mathrm{b}^{\mathrm{2}} }{\mathrm{a}^{\mathrm{2}} }\:−\:\frac{\mathrm{d}^{\mathrm{2}} }{\mathrm{c}^{\mathrm{2}} } \\ $$

Answered by MJS last updated on 13/Oct/19

α+β=(1/α)+(1/β) ⇒ β=−α ∨ β=(1/α) α^2 −β^2 =(1/α^2 )−(1/β^2 ) { ((β=−α ⇒ true for α≠0)),((β=(1/α) ⇒ true only for α=±1)) :} ⇒ question wrong? α^2 −β^2 =(1/β^2 )−(1/α^2 ) is always true for α≠0∧(β=−α∨β=(1/α))

$$\alpha+\beta=\frac{\mathrm{1}}{\alpha}+\frac{\mathrm{1}}{\beta} \\ $$$$\Rightarrow\:\beta=−\alpha\:\vee\:\beta=\frac{\mathrm{1}}{\alpha} \\ $$$$\alpha^{\mathrm{2}} −\beta^{\mathrm{2}} =\frac{\mathrm{1}}{\alpha^{\mathrm{2}} }−\frac{\mathrm{1}}{\beta^{\mathrm{2}} }\:\begin{cases}{\beta=−\alpha\:\Rightarrow\:\mathrm{true}\:\mathrm{for}\:\alpha\neq\mathrm{0}}\\{\beta=\frac{\mathrm{1}}{\alpha}\:\Rightarrow\:\mathrm{true}\:\mathrm{only}\:\mathrm{for}\:\alpha=\pm\mathrm{1}}\end{cases} \\ $$$$\Rightarrow\:\mathrm{question}\:\mathrm{wrong}?\:\alpha^{\mathrm{2}} −\beta^{\mathrm{2}} =\frac{\mathrm{1}}{\beta^{\mathrm{2}} }−\frac{\mathrm{1}}{\alpha^{\mathrm{2}} }\:\mathrm{is}\:\mathrm{always} \\ $$$$\mathrm{true}\:\mathrm{for}\:\alpha\neq\mathrm{0}\wedge\left(\beta=−\alpha\vee\beta=\frac{\mathrm{1}}{\alpha}\right) \\ $$

Commented by $@ty@m123 last updated on 13/Oct/19

Thanks for this new(for me) concept. But then where is the mistake in my soution. I am repeatedly cheking but failed to find.

$${Thanks}\:{for}\:{this}\:{new}\left({for}\:{me}\right)\:{concept}. \\ $$$${But}\:{then}\:{where}\:{is}\:{the}\:{mistake} \\ $$$${in}\:{my}\:{soution}. \\ $$$${I}\:{am}\:{repeatedly}\:{cheking} \\ $$$${but}\:{failed}\:{to}\:{find}. \\ $$

Commented by MJS last updated on 13/Oct/19

you have made no mistake and your solution is the same as mine: the question is “wrong”

$$\mathrm{you}\:\mathrm{have}\:\mathrm{made}\:\mathrm{no}\:\mathrm{mistake}\:\mathrm{and}\:\mathrm{your}\:\mathrm{solution} \\ $$$$\mathrm{is}\:\mathrm{the}\:\mathrm{same}\:\mathrm{as}\:\mathrm{mine}:\:\mathrm{the}\:\mathrm{question}\:\mathrm{is}\:``\mathrm{wrong}'' \\ $$

Commented by TawaTawa last updated on 14/Oct/19

God bless you sir

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

Answered by JDamian last updated on 13/Oct/19

if c turns into −c, then (a/b) − (c/d) = (b/a) − (d/c) ((a/b) + (c/d))×((a/b) − (c/d)) = ((b/a) + (d/c)) × ((b/a) − (d/c)) (a^2 /b^2 ) − (c^2 /d^2 ) = (b^2 /a^2 ) − (d^2 /c^2 )

$${if}\:\:\:\boldsymbol{{c}}\:\:\:{turns}\:{into}\:\:\:−\boldsymbol{{c}},\:\:{then} \\ $$$$\frac{\mathrm{a}}{\mathrm{b}}\:−\:\frac{\mathrm{c}}{\mathrm{d}}\:\:=\:\:\frac{\mathrm{b}}{\mathrm{a}}\:−\:\frac{\mathrm{d}}{\mathrm{c}} \\ $$$$ \\ $$$$\left(\frac{\mathrm{a}}{\mathrm{b}}\:+\:\frac{\mathrm{c}}{\mathrm{d}}\right)×\left(\frac{\mathrm{a}}{\mathrm{b}}\:−\:\frac{\mathrm{c}}{\mathrm{d}}\right)\:\:=\:\:\left(\frac{\mathrm{b}}{\mathrm{a}}\:+\:\frac{\mathrm{d}}{\mathrm{c}}\right)\:\:×\:\:\left(\frac{\mathrm{b}}{\mathrm{a}}\:−\:\frac{\mathrm{d}}{\mathrm{c}}\right) \\ $$$$ \\ $$$$\frac{\mathrm{a}^{\mathrm{2}} }{\mathrm{b}^{\mathrm{2}} }\:−\:\frac{\mathrm{c}^{\mathrm{2}} }{\mathrm{d}^{\mathrm{2}} }\:\:=\:\:\frac{\mathrm{b}^{\mathrm{2}} }{\mathrm{a}^{\mathrm{2}} }\:−\:\frac{\mathrm{d}^{\mathrm{2}} }{\mathrm{c}^{\mathrm{2}} } \\ $$

Commented by $@ty@m123 last updated on 13/Oct/19

I have a doubt on this process. If it would be something like: a+b=c Would you still write “if c turns into −c, then”

$${I}\:{have}\:{a}\:{doubt}\:{on}\:{this}\:{process}. \\ $$$${If}\:{it}\:{would}\:{be}\:{something}\:{like}: \\ $$$${a}+{b}={c} \\ $$$${Would}\:{you}\:{still}\:{write} \\ $$$$``{if}\:\:\:\boldsymbol{{c}}\:\:\:{turns}\:{into}\:\:\:−\boldsymbol{{c}},\:\:{then}'' \\ $$

Commented by TawaTawa last updated on 14/Oct/19

God bless you sir

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

Answered by $@ty@m123 last updated on 13/Oct/19

Solution: Given: (a/b) + (c/d) = (b/a) + (d/c) ⇒((ad+bc)/(bd))=((bc+ad)/(ac)) ⇒bd=ac .....(1) Now, LHS= (a^2 /b^2 ) − (c^2 /d^2 ) =((a^2 d^2 −b^2 c^2 )/(b^2 d^2 )) =((a^2 d^2 −b^2 c^2 )/(a^2 c^2 )) =(d^2 /c^2 )−(b^2 /a^2 ) =RHS pl. check your question.

$${Solution}: \\ $$$$\mathrm{Given}:\:\:\:\frac{\mathrm{a}}{\mathrm{b}}\:+\:\frac{\mathrm{c}}{\mathrm{d}}\:\:=\:\:\frac{\mathrm{b}}{\mathrm{a}}\:+\:\frac{\mathrm{d}}{\mathrm{c}} \\ $$$$\Rightarrow\frac{{ad}+{bc}}{{bd}}=\frac{{bc}+{ad}}{{ac}} \\ $$$$\Rightarrow{bd}={ac}\:.....\left(\mathrm{1}\right) \\ $$$${Now}, \\ $$$${LHS}=\:\frac{\mathrm{a}^{\mathrm{2}} }{\mathrm{b}^{\mathrm{2}} }\:−\:\frac{\mathrm{c}^{\mathrm{2}} }{\mathrm{d}^{\mathrm{2}} } \\ $$$$=\frac{{a}^{\mathrm{2}} {d}^{\mathrm{2}} −{b}^{\mathrm{2}} {c}^{\mathrm{2}} }{{b}^{\mathrm{2}} {d}^{\mathrm{2}} } \\ $$$$=\frac{{a}^{\mathrm{2}} {d}^{\mathrm{2}} −{b}^{\mathrm{2}} {c}^{\mathrm{2}} }{{a}^{\mathrm{2}} {c}^{\mathrm{2}} } \\ $$$$=\frac{{d}^{\mathrm{2}} }{{c}^{\mathrm{2}} }−\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} } \\ $$$$={RHS} \\ $$$${pl}.\:{check}\:{your}\:{question}. \\ $$

Commented by TawaTawa last updated on 14/Oct/19

God bless you sir

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

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