I- 6 x2-25 x-24-0II- 12 y 2-13 y -3-0A- x-y B- x - yC- x-y D- x - yE- Relationship between x and y cannot be determined

The correct option is C x lt; yI. 6x^2 + 25x + 24 = 0 D =√(b^2 4ac) D =√(625 4 × 24 × 6) =√(49)=7 x_1 = b+7/12= 25+7/12= 18/12= 3/2 x_2 = b 7/12= 25 ...

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Quantitative Aptitude

Equations

I. 6 x2+25 x+...

Question

I. $6 x^{2} + 25 x + 24 = 0$
II. $12 y^{2} + 13 y + 3 = 0$

x > y

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x \geq y

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x < y

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x \leq y

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Relationship between x and y cannot be determined

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Solution

The correct option is C x < y
I. $6 x^{2}$ + 25x + 24 = 0
$D = \sqrt{b^{2} - 4 a c} D = \sqrt{625 - 4 \times 24 \times 6} = \sqrt{49} = 7 x_{1} = \frac{- b + 7}{12} = \frac{- 25 + 7}{12} = \frac{- 18}{12} = - \frac{3}{2} x_{2} = \frac{- b - 7}{12} = \frac{- 25 - 7}{12} = \frac{- 32}{12} = - \frac{8}{3} x = \frac{- 3}{2}, \frac{- 8}{3}$
II. $12 y^{2}$ + 13y + 3 = 0
$y_{1} = \frac{- 13 + \sqrt{169 - 144}}{24} = \frac{- 13 + 5}{24} = \frac{- 8}{24} = \frac{- 1}{3} y_{2} = \frac{- 13 - \sqrt{169 - 144}}{24} = \frac{- 18}{24} = \frac{- 3}{4} y = \frac{- 1}{3}, \frac{- 3}{4} \Rightarrow x < y$

Suggest Corrections

I. 6x2+25x+24=0 II. 12y2+13y+3=0

I. $6 x^{2} + 25 x + 24 = 0$
II. $12 y^{2} + 13 y + 3 = 0$