Question

(x+y)=15
xy=54
Find the value of:
x^2-y^2

Answer 1

$xy = 54$
$(x + y) = 15$
$(x + y) {}^{2} = a {}^{2} + b {}^{2} + 2ab$
$(x + y) {}^{2} = (a - b) {}^{2} + 4ab$
$15 {}^{2} = (a - b) {}^{2} - 4 \times 54$
$(a - b) {}^{2} = 225 + 216$
$(a - b) = \sqrt{441}$
$(a - b) = 21$
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Answer 2

Answer :

$given \: = > \\ \\ \binom{(x + y) = 15}{xy = 54} \\ \\ solve \: the \: equation \: for \: x \\ \\ \binom{x = 15 - y}{xy = 54} \\ \\ substitute \: the \: value \: of \: x \: in \: the \\ equation \: xy = 54 \\ \\( 15 - y)y = 54 \\ \\ solve \: for \: y \\ \\ 15y - {y}^{2} = 54 \\ \\ 15y - {y}^{2} - 54 = 0 \\ \\ - y {}^{2} + 15y - 54 = 0 \\ \\ change \: the \: signs \\ \\ y {}^{2} - 15y + 54 = 0$

$solve \: the \: quadratic \: equation \\ \\ ax {}^{2} + bx + c = 0 \\ \\ by \: using \\ \\ x = \frac{ - b + \sqrt{ {b}^{2} - 4ac } }{2a} \\ \\ therefore \\ \\ y = \frac{ - ( - 15) + \sqrt{( - 15 {)}^{2} - 4 \times 1 \times 54} }{2 \times 1}$

$y = \frac{15 + \sqrt{225 - 4 \times 54} }{2 \times 1} \\ \\ y = \frac{15 + \sqrt{225 - 216} }{2} \\ \\ y = \frac{15 + \sqrt{9} }{2} \\ \\ y = \frac{15 + 3}{2} \\ \\ y = \frac{18}{2} = 9$

$substitute \: the \: value \: of \: y \: \\ into \: the \: equation \: x = 15 - y \\ \\ x = 15 - 9 = 6$


$now \\ \\ x {}^{2} - y {}^{2} = > \\ {6}^{2} - {9}^{2} \\ \\ = > 36 - 81 \\ \\ = > \: - 45$


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