Question

Find the zeroes of the following pair of equation :

x + y = 3

x² + y² = 29

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Answer 1

x + y = 3 ___ [1] ( Given )


x² + y² = 29 ( Given )


$<b>( a + b )² = a² + b² + 2ab</b>$


Here, a = x, b = y


⇒ ( 3 )² = 29 + 2xy


⇒ 9 = 29 + 2xy


⇒ 9 - 29 = 2xy


⇒ - 20 = 2xy


⇒ xy = - 20 / 2


⇒ xy = - 10


⇒ x = - 10 / y


Put this value in [1], we get


⇒ ( - 10 / y ) + y = 3


⇒ ( - 10 + y² ) / y = 3


⇒ y² - 3y - 10 = 0


By Middle Term Factorisation


⇒ y² - 5y + 2y - 10 = 0


⇒ y ( y - 5 ) + 2 ( y - 5 ) = 0


⇒ ( y + 2 ) ( y - 5 ) = 0


Using Zero Product Rule


⇒ ( y + 2 ) = 0 and ( y - 5 ) = 0


⇒ y = - 2, 5


Now,


$<u>Case I : when y = - 2</u>$


x + y = 3


⇒ x - 2 = 3


⇒ x = 3 + 2


⇒ x = 5


$<u>Case II : when y = 5</u>$


x + y = 3


⇒ x + 5 = 3


⇒ x = 3 - 5


⇒ x = - 2



Hence, $<b>The zeroes are - 2 and 5.</b>$

Answer 2

x=-2

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