I- 4 x2-13 x-12-0II- y 2-7 y -60-0A- x-y B- x - yC- x - yD- x - yE- x - y or relationship cannot be established

The correct option is E x = y or relationship cannot be established(x 4)(4x + 3) = 0 x = 4, 3/4 (x + 5)(x 12) = 0 y = 12, 5 Option e is the right answer. ...

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Quantitative Aptitude

Solving Inequalities

I. 4 x2-13 x-...

Question

I. $4 x^{2} - 13 x - 12 = 0$
II. $y^{2} - 7 y - 60 = 0$

x > y

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x < y

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x \geq y

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x \leq y

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x = y

or relationship cannot be established

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Solution

The correct option is E $x = y$ or relationship cannot be established
$(x - 4) (4 x + 3) = 0$
$x = 4, \frac{- 3}{4}$
$(x + 5) (x - 12) = 0$
$y = 12, - 5$
Option e is the right answer.

Suggest Corrections

I. 4x2−13x−12=0 II. y2−7y−60=0

The correct option is E x=y or relationship cannot be established(x−4)(4x+3)=0 x=4,−34 (x+5)(x−12)=0 y=12,−5 Option e is the right answer.

I. $4 x^{2} - 13 x - 12 = 0$
II. $y^{2} - 7 y - 60 = 0$

The correct option is E $x = y$ or relationship cannot be established
$(x - 4) (4 x + 3) = 0$
$x = 4, \frac{- 3}{4}$
$(x + 5) (x - 12) = 0$
$y = 12, - 5$
Option e is the right answer.