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 cannot be established

The correct option is E x = y or cannot be established(x 4)(4x+3)=0 x=4, 3/4 (x+5)(x 12)=0 y=12, 5 ...

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Maxima/Minima of a Quadratic Equation

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 \leq y

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x \geq y

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x = y

or cannot be established

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