Math, asked by haider06, 6 months ago

√5x² + 7x + 2 - √4x² + 7x +18 = x-4

Answers

Answered by Flaunt
30

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√5x² + 7x + 2 - √4x² + 7x +18 = x-4

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 =  >  \sqrt{5}  {x}^{2}  + 7x + 2 -  \sqrt{4}  {x}^{2}  + 7x + 18 = x - 4

Squaring both sides:-

 =  >   {( \sqrt{5}  {x}^{2}) }^{2}  +  {(7x)}^{2}  +  {(2)}^{2}  -  {( \sqrt{4} {x}^{2} ) }^{2}  +  {(7x)}^{2}  +  {(18)}^{2}  =  {x}^{2}  -  {4}^{2}

 =  > 5x  + 7{x}^{2}  + 4 - 4x + 7 {x}^{2}  + 324 =  {x}^{2}  - 16

 =  > 7 {x}^{2}  + 7 {x}^{2}  + 5x - 4x + 4 + 324 =  {x}^{2} - 16

 =  > 14 {x}^{2}  + x + 328 =  {x}^{2}  - 16

 =  > 13 {x}^{2}  + x + 344 = 0

Here ,we use quadratic formula to find out the solution:-

here,a=13;b=1 & c=344

 =  > x =  \frac{ -  b± \sqrt{ {b}^{2} - 4ac } }{2a}

 =  > x =  \frac{ - 1± \sqrt{1 - 4 \times 344} }{2}

 =  > x =  \frac{ - 1± \sqrt{1 - 1376} }{2}

 =  > x =  \frac{ - 1± \sqrt{ - 1375} }{2}

 =  > x =  \frac{ - 1±5 \sqrt{55i} }{2}

=>Here,we see that negative value inside root so in order to replace negative sign from inside root we put 'i' ( iota) in place of negative sign.

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