I- - x - 2-10 x -24-011II- - 4 y - 2-17 y -18-0 A- x yE- Relationship between x and y cannot be established

The correct option is A x lt; yEquation I x^2+10x+24=0 ⇒ x^2+6x+4x+24=0 ⇒ x(x+6)+4(x+6)=0 ⇒ (x+4)(x+6)=0 ∴ x= 4, 6. Equation II 4y^2 17y+18=0 ⇒ 4y^2 8y 9y+ ...

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Byju's Answer

Standard XII

Mathematics

Derivative of Standard Functions

I. ∼ x ∧ 2+10...

Question

$I . x^{2} + 10 x + 24 = 0 I I . 4 y^{2} - 17 y + 18 = 0$

x \leq y

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x \geq y

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Relationship between x and y cannot be established

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x > y

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x < y

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Solution

The correct option is E $x < y$
Equation I
$x^{2} + 10 x + 24 = 0$
$\Rightarrow x^{2} + 6 x + 4 x + 24 = 0$
$\Rightarrow x (x + 6) + 4 (x + 6) = 0$
$\Rightarrow (x + 4) (x + 6) = 0$
$∴ x = - 4, - 6$ .
Equation II
$4 y^{2} - 17 y + 18 = 0$
$\Rightarrow 4 y^{2} - 8 y - 9 y + 18 = 0$
$\Rightarrow 4 y (y - 2) - 9 (y - 2) = 0$
$\Rightarrow (4 y - 2) (y - 2) = 0$
$∴ y = \frac{9}{4}, 2$
Hence, x < y

Suggest Corrections

I. x2+10x+24=0II. 4y2−17y+18=0

The correct option is E x<yEquation I x2+10x+24=0 ⇒x2+6x+4x+24=0 ⇒x(x+6)+4(x+6)=0 ⇒(x+4)(x+6)=0 ∴x=−4,−6. Equation II 4y2−17y+18=0 ⇒4y2−8y−9y+18=0 ⇒4y(y−2)−9(y−2)=0 ⇒(4y−2)(y−2)=0 ∴y=94,2 Hence, x < y

$I . x^{2} + 10 x + 24 = 0 I I . 4 y^{2} - 17 y + 18 = 0$