Math, asked by kshitijvarshney, 1 year ago

plz solve this ques }pppppppppppppppppppp

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kshitijvarshney: plz solve it

Answers

Answered by Eustacia
0

(  \: {a}^{2}  +  {b}^{2} \:  ) {x}^{2}  - 2b( \: a + c \: )x \:  +  \: ( \:  {b}^{2}  +  {c}^{2}  \: ) = 0 \\  \\ Roots \:  \:  are  \:  \: real  \:  \: and \:  \:  equal \:  ,  \\ therefore  \:  \: D=0 \\  \\ D  = \:    (- 2b( \: a + c \: )) {}^{2}  - 4(  \: {a}^{2}  +  {b}^{2} \:  )( \:  {b}^{2}  +  {c}^{2}  \: ) = 0 \\  \\  {b}^{2}  \: ( \: a {}^{2}  + 2ac \:  +  {c }^{2} \:  ) \:  =   {a}^{2}  {b}^{2}  +  {a}^{2}  {c}^{2}  +  {b}^{2}  {c}^{2}  +  {b}^{4}  \\  \\ ( \: ac \: ) {}^{2}  - 2ac( {b}^{2} ) +  ({b}^{2} ) {}^{2}  = 0 \\ \\  \:  \:  \: ( \:  ac \:  -  \:  {b}^{2}  \: ) {}^{2}  \:  =  \: 0 \\  \\  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \large \boxed { \: b {}^{2}  \:  =  \: ac } \\  \\ a \:  \:  \:  ,   \: \:  \: b  \: \:  ,  \:  \: c  \:  \:  \: are  \:  \: in \:  \:  \:  G.P.
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