I- 10 x 2-33 x -27-0II- 5 y 2-19 y -18-0A- x-yB- x-y C- x - yD- x - yE- Relationship between x and y cannot be determined

The correct option is C x ≥ yI. 10x^2 + 33x + 27 = 0 x_1 = 33+√(b^2 4ac)/2a= 33+√(1089 4 × 10 × 27)/20 x_2 = 33 √(b^2 4ac)/2a x_2 = 33 √(1089 1080)/20 x_1 ...

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Quantitative Aptitude

Quadratic Equations

I. 10 x 2+33 ...

Question

I. $10 x^{2} + 33 x + 27 = 0$
II. $5 y^{2} + 19 y + 18 = 0$

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 B $x \geq y$
I. $10 x^{2}$ + 33x + 27 = 0
$x_{1} = \frac{- 33 + \sqrt{b^{2} - 4 a c}}{2 a} = \frac{- 33 + \sqrt{1089 - 4 \times 10 \times 27}}{20} x_{2} = \frac{- 33 - \sqrt{b^{2} - 4 a c}}{2 a} x_{2} = \frac{- 33 - \sqrt{1089 - 1080}}{20} x_{1} = \frac{- 33 + 3}{20}, x_{2} = \frac{- 33 - 3}{20} x_{1} = \frac{- 30}{20}, x_{2} = \frac{- 36}{20} = \frac{- 9}{5}, x = \frac{- 3}{2}, \frac{- 9}{5}$
II. $5 y^{2}$ + 19 y + 18 = 0
$y_{1} = \frac{- 19 + \sqrt{361 - 4 \times 18 \times 5}}{10} y_{2} = \frac{- 19 - \sqrt{361 - 360}}{10} y_{1} = \frac{- 19 + 1}{10} = \frac{- 18}{10} = \frac{- 9}{5} y_{2} = \frac{- 19 - 1}{10} = \frac{- 20}{10} = - 2 y = \frac{- 9}{5}, - 2 \Rightarrow x \geq y$

Suggest Corrections

I. 10x2+33x+27=0 II. 5y2+19y+18=0

I. $10 x^{2} + 33 x + 27 = 0$
II. $5 y^{2} + 19 y + 18 = 0$