In the following question, two equations numbered I and II are given. You have to solve both the equations and give answer if
(I) $x^{2} + 11 x + 28 = 0$ and (II) $y^{2} + 15 y + 56 = 0$

x > y

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x \geq y

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x < y

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x \leq y

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Solution

The correct option is A $x \geq y$
$\Rightarrow x^{2} + 11 x + 28 = 0$
$(x + 7) (x + 4) = 0$
$x = - 7$ and $x = - 4$
$\Rightarrow y^{2} + 15 y + 56 = 0$
$(y + 7) (y + 8) = 0$
$y = - 7, - 8$
Hence, $X \geq Y$ .

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In the following question, two equations numbered I and II are given. You have to solve both the equations and give answer if(I) x2+11x+28=0 and (II) y2+15y+56=0

The correct option is A x≥y⇒x2+11x+28=0(x+7)(x+4)=0x=−7 and x=−4⇒y2+15y+56=0(y+7)(y+8)=0y=−7,−8Hence, X≥Y.

In the following question, two equations numbered I and II are given. You have to solve both the equations and give answer if
(I) $x^{2} + 11 x + 28 = 0$ and (II) $y^{2} + 15 y + 56 = 0$

The correct option is A $x \geq y$
$\Rightarrow x^{2} + 11 x + 28 = 0$
$(x + 7) (x + 4) = 0$
$x = - 7$ and $x = - 4$
$\Rightarrow y^{2} + 15 y + 56 = 0$
$(y + 7) (y + 8) = 0$
$y = - 7, - 8$
Hence, $X \geq Y$ .