In the following question, two equations numbered I and II are given. Solve both the equations and mark the appropriate option:

$I . 2 x^{2} + 40 = 18 x$
$I I . y^{2} = 13 y + 42$

x = y

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x > y

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x < y

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Solution

The correct option is C x < y
$F r o m I, 2 x^{2} + 40 = 18 x 2 x^{2} + 40 - 19 x = 0 x^{2} - 9 x - 20 = 0 (x - 4) (x - 5) = 0 x = 4 o r 5 \dots (1) F r o m I I, y^{2} = 13 y + 42 y^{2} - 13 y - 42 = 0 (y - 7) (y - 6) = 0 y = 7 o r y = 6 \dots (2)$
From (1) and (2), we have $y > x$

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