Menu Close

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-




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

Leave a Reply

Your email address will not be published. Required fields are marked *