I- x2-5 x-6-0II- y 2-7 y -12-0A- x - yB- x-y C- x - yD- x-y or no relation can be established between x and yE- x-y

The correct option is A x ≥ yEquation I x^2+5x+6=0 ⇒ x^2+2x+3x+6=0 ⇒ x(x+2)+3(x+2)=0 ⇒ (x+3)(x+2)=0 ∴ x= 3 or 2 Equation II y^2+7y+12=0 ⇒ y^2+3y+4y+12=0 ⇒ y( ...

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Byju's Answer

Standard XII

Mathematics

Derivative of Standard Functions

I. x2+5 x+6=0...

Question

$I . x^{2} + 5 x + 6 = 0$
$I I . y^{2} + 7 y + 12 = 0$

\geq

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x > y

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\leq

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x < y

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x = y or no relation can be established between x and y

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Solution

The correct option is A x $\geq$ y
Equation I
$x^{2} + 5 x + 6 = 0$
$\Rightarrow x^{2} + 2 x + 3 x + 6 = 0$
$\Rightarrow x (x + 2) + 3 (x + 2) = 0$
$\Rightarrow (x + 3) (x + 2) = 0$
$∴ x = - 3 o r - 2$
Equation II
$y^{2} + 7 y + 12 = 0$
$\Rightarrow y^{2} + 3 y + 4 y + 12 = 0$
$\Rightarrow y (y + 3) + 4 (y + 3) = 0$
$\Rightarrow (y + 4) (y + 3) = 0$
$∴ y = - 3 o r - 4$
Hence, $x \geq y$

Suggest Corrections

I. x2+5x+6=0 II. y2+7y+12=0

The correct option is A x ≥ yEquation I x2+5x+6=0 ⇒x2+2x+3x+6=0 ⇒x(x+2)+3(x+2)=0 ⇒(x+3)(x+2)=0 ∴x=−3 or −2 Equation II y2+7y+12=0 ⇒y2+3y+4y+12=0 ⇒y(y+3)+4(y+3)=0 ⇒(y+4)(y+3)=0 ∴y=−3 or −4 Hence, x≥y

$I . x^{2} + 5 x + 6 = 0$
$I I . y^{2} + 7 y + 12 = 0$