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$6x^2 +5x+1=0$
$15y^2 +8y+1=0$

  1. $x >y$
  2. $x \leq y$
  3. $x < y$
  4. $x \geq y$
  5. Relationship between X and Y cannot be established

 

please provide explanation

 

asked in Logical Reasoning by (7.6k points)   | 22 views

1 Answer

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We can find the roots of a quadratic equation with the quadratic formula

$x =\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$

 

For 1st  equation

$x =\frac{-5 \pm \sqrt{25-24}}{12}$

$x =\frac{-5 \pm 1}{12}$

$x =-0.5,-0.33$

 

For 2nd equation

$15y^{2}+8y+1$=0

$15y^{2}+3y+5y+1$

y=$y=- \frac{1}{5} , -\frac{1}{3}$ 

(We can also find the roots by same method used for 1st equation.)

$y=- 0.2 , -0.33$
 

Here,

$- 0.2 \ , \ -0.33 \ \geq -0.5 \ , \ -0.33$

Hence relation between x and y = $y \geq x$

 

Hence,Option $(B) x\leq y .$ is the correct choice.

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