Question

√5x² + 7x + 2 - √4x² + 7x +18 = x-4​

Answer 1

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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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