I- - 6 x- 2-29 x-35-0 1II- - 3 y- 2-11 y-10-0 A- x-y or relationship cannot be establishedB- x - yC- x - yD- x-y E- x - y

The correct option is D x gt; yEquation I 6x^2 29x+35=0 ⇒ 6x^2 15x 14x+35=0 ⇒ 3x(2x 5) 7(2x 5)=0 ⇒ (3x 7)(2x 5)=0 ⇒ x=7/3,x=5/2 Equation II 3y^2 11y+10=0 ⇒ ...

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

Byju's Answer

Standard XII

Mathematics

Derivative of Standard Functions

I. ∼ 6 x∧ 2-2...

Question

$I . 6 x^{2} - 29 x + 35 = 0 I I . 3 y^{2} - 11 y + 10 = 0$

x \geq y

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

x = y or relationship cannot be established

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

x \leq 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 < y

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

Open in App

Solution

The correct option is D $x > y$
Equation I
$6 x^{2} - 29 x + 35 = 0$
$\Rightarrow 6 x^{2} - 15 x - 14 x + 35 = 0$
$\Rightarrow 3 x (2 x - 5) - 7 (2 x - 5) = 0$
$\Rightarrow (3 x - 7) (2 x - 5) = 0$
$\Rightarrow x = \frac{7}{3}, x = \frac{5}{2}$
Equation II
$3 y^{2} - 11 y + 10 = 0$
$\Rightarrow 3 y^{2} - 6 y - 5 y + 10 = 0$
$\Rightarrow 3 y (y - 2) - 5 (y - 2) = 0$
$\Rightarrow (3 y - 5) (y - 2) = 0$
$\Rightarrow y = \frac{5}{3}, y = 2$
Hence, x > y

Suggest Corrections

I. 6x2−29x+35=0II. 3y2−11y+10=0

The correct option is D x>yEquation I 6x2−29x+35=0 ⇒6x2−15x−14x+35=0 ⇒3x(2x−5)−7(2x−5)=0 ⇒(3x−7)(2x−5)=0 ⇒x=73,x=52 Equation II 3y2−11y+10=0 ⇒3y2−6y−5y+10=0 ⇒3y(y−2)−5(y−2)=0 ⇒(3y−5)(y−2)=0 ⇒y=53,y=2 Hence, x > y

$I . 6 x^{2} - 29 x + 35 = 0 I I . 3 y^{2} - 11 y + 10 = 0$