I- x2-4-0II- y 2-6 y -9-0A- x-y or the relationship cannot be established-B- If x-yC- If x - yD- If x-yE- If x - y

The correct option is B If x gt; yEquation I x^2 4=0 ⇒ (x+2)(x 2)=0 ∴ x = 2 or 2 Equation II y^2+6y+9=0 ⇒ y^2 + 3y+3y+9=0 ⇒ y(y+3)+3(y+3)=0 ⇒ (y+3)(y+3) ...

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Byju's Answer

Standard XII

Mathematics

Derivative of Standard Functions

I. x2-4=0II. ...

Question

I. $x^{2} - 4 = 0$
II. $y^{2} + 6 y + 9 = 0$

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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x = y

or the relationship cannot be established.

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Solution

The correct option is A If $x > y$
Equation I
$x^{2} - 4 = 0$
$\Rightarrow (x + 2) (x - 2) = 0$
$∴ x = 2$ or $- 2$

Equation II
$y^{2} + 6 y + 9 = 0$
$\Rightarrow y^{2} + 3 y + 3 y + 9 = 0$
$\Rightarrow y (y + 3) + 3 (y + 3) = 0$
$\Rightarrow (y + 3) (y + 3) = 0$
$y = - 3, - 3$
Hence, $x > y$

Suggest Corrections

I. x2−4=0 II. y2+6y+9=0

The correct option is A If x>yEquation I x2−4=0 ⇒(x+2)(x−2)=0 ∴x=2 or −2 Equation II y2+6y+9=0 ⇒y2+3y+3y+9=0 ⇒y(y+3)+3(y+3)=0 ⇒(y+3)(y+3)=0 y=−3,−3 Hence, x>y

I. $x^{2} - 4 = 0$
II. $y^{2} + 6 y + 9 = 0$

The correct option is A If $x > y$
Equation I
$x^{2} - 4 = 0$
$\Rightarrow (x + 2) (x - 2) = 0$
$∴ x = 2$ or $- 2$

Equation II
$y^{2} + 6 y + 9 = 0$
$\Rightarrow y^{2} + 3 y + 3 y + 9 = 0$
$\Rightarrow y (y + 3) + 3 (y + 3) = 0$
$\Rightarrow (y + 3) (y + 3) = 0$
$y = - 3, - 3$
Hence, $x > y$