Question

plz solve this problem ​

Answer 1

hey mate here ur answer




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Answer 2

Roots are equal. So discriminant is equal.

                       

$(-2(ac+bd))^2-(4(a^2+b^2)(c^2+d^2))=0 \\ \\ (-2ac-2bd)^2-(4((ac)^2+(ad)^2+(bc)^2+(bd)^2))=0 \\ \\ (4(ac)^2+8abcd+4(bd)^2)-(4(ac)^2+4(ad)^2+4(bc)^2+4(bd)^2)=0 \\ \\ 4(ac)^2+8abcd+4(bd)^2-4(ac)^2-4(ad)^2-4(bc)^2-4(bd)^2=0 \\ \\ 8abcd-4(ad)^2-4(bc)^2=0 \\ \\ 2abcd-(ad)^2-(bc)^2=0 \\ \\ -1 \times (2abcd-(ad)^2-(bc)^2)=-1\times0$

         

$-2abcd+(ad)^2+(bc)^2=0 \\ \\ (ad)^2-2adbc+(bc)^2=0 \\ \\ (ad-bc)^2=0 \\ \\ ad-bc=0 \\ \\ ad=bc \\ \\ a=\frac{bc}{d} \\ \\ \frac{a}{b}=\frac{c}{d}$

           

Hence proved!!!

     

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