I- 2 X 2-5 X -3-0II- Y 2-9 Y -14-0A- Relationship between X and Y cannot be establishedB- X - YC- X-YD- X - YE- X-Y

The correct option is C X gt;YEquation I 2X^2+5X+3=0 ⇒ 2X^2+2X+3X+3=0 ⇒ 2X(X+1)+3(X+1)=0 ⇒ (X+1)(2X+3)=0 ∴ X= 3/2 or 1 Equation II Y^2+9Y+14=0 ⇒ Y^2+7Y+2Y+1 ...

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Byju's Answer

Standard XII

Mathematics

Derivative of Standard Functions

I. 2 X 2+5 X ...

Question

$I . 2 X^{2} + 5 X + 3 = 0$
$I I . Y^{2} + 9 Y + 14 = 0$

X \geq Y

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X > Y

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X \leq Y

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X < Y

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Relationship between X and Y cannot be established

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Solution

The correct option is B $X > Y$
Equation I
$2 X^{2} + 5 X + 3 = 0$
$\Rightarrow 2 X^{2} + 2 X + 3 X + 3 = 0$
$\Rightarrow 2 X (X + 1) + 3 (X + 1) = 0$
$\Rightarrow (X + 1) (2 X + 3) = 0$
$∴ X = - \frac{3}{2} o r - 1$
Equation II
$Y^{2} + 9 Y + 14 = 0$
$\Rightarrow Y^{2} + 7 Y + 2 Y + 14 = 0$
$\Rightarrow Y (Y + 7) + 2 (Y + 7) = 0$
$\Rightarrow (Y + 2) (Y + 7) = 0$
$∴ Y = - 2, Y = - 7$
Hence, $X > Y$

Suggest Corrections

I. 2X2+5X+3=0 II. Y2+9Y+14=0

The correct option is B X>YEquation I 2X2+5X+3=0 ⇒2X2+2X+3X+3=0 ⇒2X(X+1)+3(X+1)=0 ⇒(X+1)(2X+3)=0 ∴X=−32 or −1 Equation II Y2+9Y+14=0 ⇒Y2+7Y+2Y+14=0 ⇒Y(Y+7)+2(Y+7)=0 ⇒(Y+2)(Y+7)=0 ∴Y=−2,Y=−7 Hence, X>Y

$I . 2 X^{2} + 5 X + 3 = 0$
$I I . Y^{2} + 9 Y + 14 = 0$