I- 2 x 2-11 x -14-0II- 4 y 2-12 y -9-0A- If xy

The correct option is A If x lt; yEquation I 2x^2+11x+14=0 ⇒ 2x^2+4x+7x+14=0 ⇒ 2x(x+2)+7(x+2)=0 ⇒ (x+2)(2x+7)=0 ∴ x= 2, 7/2 Equation II 4y^2 +12y+9=0 ⇒ 4y ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Derivative of Standard Functions

I. 2 x 2+11 x...

Question

I. $2 x^{2} + 11 x + 14 = 0$
II. $4 y^{2} + 12 y + 9 = 0$

x > y

No worries! We‘ve got your back. Try BYJU‘S free classes today!

x \geq y

No worries! We‘ve got your back. Try BYJU‘S free classes today!

x < y

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

x \leq y

No worries! We‘ve got your back. Try BYJU‘S free classes today!

x = y

or the relationship cannot be established.

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C If $x < y$
Equation I
$2 x^{2} + 11 x + 14 = 0$
$\Rightarrow 2 x^{2} + 4 x + 7 x + 14 = 0$
$\Rightarrow 2 x (x + 2) + 7 (x + 2) = 0$
$\Rightarrow (x + 2) (2 x + 7) = 0$
$∴ x = - 2, - \frac{7}{2}$

Equation II
$4 y^{2} + 12 y + 9 = 0$
$\Rightarrow 4 y^{2} + 6 y + 6 y + 9 = 0$
$\Rightarrow 2 y (2 y + 3) + 3 (2 y + 3) = 0$
$\Rightarrow (2 y + 3)^{2} = 0$
$∴ y = - \frac{3}{2}, - \frac{3}{2}$
Hence, $x < y$

Suggest Corrections

I. 2x2+11x+14=0 II. 4y2+12y+9=0

The correct option is C If x<yEquation I 2x2+11x+14=0 ⇒2x2+4x+7x+14=0 ⇒2x(x+2)+7(x+2)=0 ⇒(x+2)(2x+7)=0 ∴x=−2,−72 Equation II 4y2+12y+9=0 ⇒4y2+6y+6y+9=0 ⇒2y(2y+3)+3(2y+3)=0 ⇒(2y+3)2=0 ∴y=−32,−32 Hence, x<y

I. $2 x^{2} + 11 x + 14 = 0$
II. $4 y^{2} + 12 y + 9 = 0$